VSTStoichiometry
Virtual Science Teac
Created on March 9, 2022
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Transcript
Mass of A to Moles of A
Moles of A to Mass of A
Moles of A to Particles of A
Stoichiometry Made Simple
Particles of A to Moles of A
Volume of Gas A to Moles of A
Moles of A to Volume of Gas A
Mass of A to Particles of A
Particles of A to Mass of A
Particles of A to Volume of Gas A
Volume of Gas A to Particles of A
Mass of Gas A to Volume of Gas A
Volume of Gas A to Mass of Gas A
Mass of A to Moles of B
Volume of A to Volume of B
Particles of A to Particles of B
Brought to you by Virtual Science Teachers
Mass of A to Mass of B
Select a green starting position on the diagram.
Use balanced equation
1 Mole = 6.02 x 1023 particles
particlesB
1 Mole = 6.02 x 1023 particles
mass A(grams)
Use Molar Mass
particlesA
volume of gas A(liters)
molesB
Use balanced equation
1 Mole = 22.4 L of gas at STP
molesA
mass B(grams)
Use Molar Mass
volume of gas B(liters)
1 Mole = 22.4 L of gas at STP
Use balanced equation
This interactive provides step-by-step explanations and examples of many different types of stoichiometry problems.The diagram below shows the possible "paths" that can be taken to get from one quantity to another.
Stoichiometry /ˌstɔɪkiˈɒmɪtri/ refers to the relationship between the quantities of reactants and products before, during, and following chemical reactions.
Copyright 2022 Virtual Science Teachers
Brought to you by Virtual Science Teachers
Mass of A to Moles of A
Moles of A to Mass of A
Moles of A to Particles of A
Stoichiometry Made Simple
What kind of problem would you like to solve?
Particles of A to Moles of A
Volume of Gas A to Moles of A
Moles of A to Volume of Gas A
Mass of A to Particles of A
Particles of A to Mass of A
Particles of A to Volume of Gas A
Volume of Gas A to Particles of A
Mass of Gas A to Volume of Gas A
Volume of Gas A to Mass of Gas A
Mass of A to Moles of B
Volume of A to Volume of B
Particles of A to Particles of B
Use balanced equation
Brought to you by Virtual Science Teachers
Mass of A to Mass of B
1 Mole = 6.02 x 1023 particles
particlesB
1 Mole = 6.02 x 1023 particles
mass A(grams)
Use Molar Mass
particlesA
volume of gas A(liters)
molesB
Use balanced equation
1 Mole = 22.4 L of gas at STP
molesA
mass B(grams)
Use Molar Mass
volume of gas B(liters)
1 Mole = 22.4 L of gas at STP
Use balanced equation
Mass of A to Moles of A
Moles of A to Mass of A
Moles of A to Particles of A
Stoichiometry Made Simple
What kind of problem would you like to solve?
Particles of A to Moles of A
Volume of Gas A to Moles of A
Moles of A to Volume of Gas A
Mass of A to Particles of A
Particles of A to Mass of A
Particles of A to Volume of Gas A
Volume of Gas A to Particles of A
Mass of Gas A to Volume of Gas A
Volume of Gas A to Mass of Gas A
Mass of A to Moles of B
Volume of A to Volume of B
Particles of A to Particles of B
Use balanced equation
Brought to you by Virtual Science Teachers
Mass of A to Mass of B
Mass of A to Volume of B
Mass of A to Particles of B
Mass of A to Moles of B
Volume of A to Moles of B
1 Mole = 6.02 x 1023 particles
Volume of A to Mass of B
particlesB
1 Mole = 6.02 x 1023 particles
mass A(grams)
Use Molar Mass
particlesA
volume of gas A(liters)
molesB
Use balanced equation
1 Mole = 22.4 L of gas at STP
molesA
mass B(grams)
Use Molar Mass
volume of gas B(liters)
1 Mole = 22.4 L of gas at STP
Use balanced equation
Mass of A to Moles of A
Moles of A to Mass of A
Moles of A to Particles of A
Stoichiometry Made Simple
Particles of A to Moles of A
Volume of Gas A to Moles of A
Moles of A to Volume of Gas A
Mass of A to Particles of A
Particles of A to Mass of A
Particles of A to Volume of Gas A
Volume of Gas A to Particles of A
Mass of Gas A to Volume of Gas A
Volume of Gas A to Mass of Gas A
Mass of A to Moles of B
Volume of A to Volume of B
Particles of A to Particles of B
Use balanced equation
Brought to you by Virtual Science Teachers
Mass of A to Mass of B
Now select an ending position on the diagram.
1 Mole = 6.02 x 1023 particles
Any path that leads to a blue circle involves the quantities of products and reactants involved in a chemical reaction.
particlesB
1 Mole = 6.02 x 1023 particles
mass A(grams)
Use Molar Mass
particlesA
volume of gas A(liters)
molesB
Use balanced equation
1 Mole = 22.4 L of gas at STP
molesA
mass B(grams)
Use Molar Mass
volume of gas B(liters)
1 Mole = 22.4 L of gas at STP
Use balanced equation
Mass of A to Moles of A
Moles of A to Mass of A
Moles of A to Particles of A
Stoichiometry Made Simple
Particles of A to Moles of A
Volume of Gas A to Moles of A
Moles of A to Volume of Gas A
Mass of A to Particles of A
Particles of A to Mass of A
Particles of A to Volume of Gas A
Volume of Gas A to Particles of A
Mass of Gas A to Volume of Gas A
Volume of Gas A to Mass of Gas A
Mass of A to Moles of B
Volume of A to Volume of B
Particles of A to Particles of B
Use balanced equation
Brought to you by Virtual Science Teachers
Mass of A to Mass of B
Now select an ending position on the diagram.
1 Mole = 6.02 x 1023 particles
particlesB
Coming Soon
Any path that leads to a blue circle involves the quantities of products and reactants involved in a chemical reaction.
1 Mole = 6.02 x 1023 particles
mass A(grams)
Use Molar Mass
particlesA
volume of gas A(liters)
molesB
Use balanced equation
1 Mole = 22.4 L of gas at STP
molesA
mass B(grams)
Use Molar Mass
volume of gas B(liters)
1 Mole = 22.4 L of gas at STP
Use balanced equation
Mass of A to Moles of A
Moles of A to Mass of A
Moles of A to Particles of A
Stoichiometry Made Simple
Particles of A to Moles of A
Volume of Gas A to Moles of A
Moles of A to Volume of Gas A
Mass of A to Particles of A
Particles of A to Mass of A
Particles of A to Volume of Gas A
Volume of Gas A to Particles of A
Mass of Gas A to Volume of Gas A
Volume of Gas A to Mass of Gas A
Mass of A to Moles of B
Volume of A to Volume of B
Particles of A to Particles of B
Use balanced equation
Brought to you by Virtual Science Teachers
Mass of A to Mass of B
Now select an ending position on the diagram.
Coming Soon
Coming Soon
Coming Soon
Any path that leads to a blue circle involves the quantities of products and reactants involved in a chemical reaction.
1 Mole = 6.02 x 1023 particles
particlesB
1 Mole = 6.02 x 1023 particles
mass A(grams)
Use Molar Mass
particlesA
volume of gas A(liters)
molesB
Use balanced equation
1 Mole = 22.4 L of gas at STP
molesA
mass B(grams)
Use Molar Mass
volume of gas B(liters)
1 Mole = 22.4 L of gas at STP
Use balanced equation
Mass of A to Moles of A
Moles of A to Mass of A
Moles of A to Particles of A
Stoichiometry Made Simple
Particles of A to Moles of A
Volume of Gas A to Moles of A
Moles of A to Volume of Gas A
Mass of A to Particles of A
Particles of A to Mass of A
Particles of A to Volume of Gas A
Volume of Gas A to Particles of A
Mass of Gas A to Volume of Gas A
Volume of Gas A to Mass of Gas A
Mass of A to Moles of B
Volume of A to Volume of B
Particles of A to Particles of B
Use balanced equation
Brought to you by Virtual Science Teachers
Mass of A to Mass of B
Now select an ending position on the diagram.
1 Mole = 6.02 x 1023 particles
particlesB
Coming Soon
Coming Soon
Coming Soon
Any path that leads to a blue circle involves the quantities of products and reactants involved in a chemical reaction.
1 Mole = 6.02 x 1023 particles
mass A(grams)
Use Molar Mass
particlesA
volume of gas A(liters)
molesB
Use balanced equation
1 Mole = 22.4 L of gas at STP
molesA
mass B(grams)
Use Molar Mass
volume of gas B(liters)
1 Mole = 22.4 L of gas at STP
Use balanced equation
Next
Use balanced equation
1 Mole = 6.02 x 1023 particles
*Note: This simulation shows the immediate water displacement when a rainbow swirl is added to water, before it dissolves.
particlesB
1 Mole = 6.02 x 1023 particles
mass A(grams)
Use Molar Mass
particlesA
volume of gas A(liters)
molesB
Use balanced equation
1 Mole = 22.4 L of gas at STP
molesA
mass B(grams)
Use Molar Mass
volume of gas B(liters)
1 Mole = 22.4 L of gas at STP
Use balanced equation
Step 2: Draw a multiplication sign and another line.
106 grams NaCl
1
x
Step 1: Put what you start with over 1.
106 grams NaCl
1
mass(grams)
106 grams NaCl = ? moles NaCl
106 grams NaCl
1
Step 3: Write the units you want to cancel out the denominator of this new fraction.
x
grams NaCl
Step 4: Write the units of your next stop in the numerator.
x
grams NaCl
mole NaCl
106 grams NaCl
1
1 mole of NaCl has a mass of 58.4 grams.
Step 5: Use the corresponding conversion factor to write in the values.
x
58.4 grams NaCl
1 mole NaCl
106 grams NaCl
1
Step 6: Cancel out units that are in both the numerator and denominator. Then, multiply the top numbers and divide by the bottom numbers to get your answer!
x
58.4 grams NaCl
1 mole NaCl
=
106 grams NaCl
1
1.82 moles NaCl
Next
Use balanced equation
1 Mole = 6.02 x 1023 particles
*Note: This simulation shows the immediate water displacement when a rainbow swirl is added to water, before it dissolves.
particlesB
1 Mole = 6.02 x 1023 particles
mass A(grams)
Use Molar Mass
particlesA
volume of gas A(liters)
molesB
Use balanced equation
1 Mole = 22.4 L of gas at STP
molesA
mass B(grams)
Use Molar Mass
volume of gas B(liters)
1 Mole = 22.4 L of gas at STP
Use balanced equation
25.3 moles H20 = ? grams H20
Step 2: Draw a multiplication sign and another line.
25.3 moles H20
1
x
Step 1: Put what you start with over 1.
25.3 moles H20
1
25.3 moles H2O
1
Step 3: Write the units you want to cancel out in the denominator of this new fraction.
x
mole H20
Step 4: Write the units of your next stop in the numerator.
x
mole H2O
grams H20
25.3 moles H2O
1
Step 5: Use the corresponding conversion factor to write in the values.
x
1 mole H2O
18.0 grams H2O
25.3 moles H20
1
1 mole of H2O has a mass of 18.0 grams.
Step 6: Cancel out units that are in both the numerator and denominator. Then, multiply the top numbers and divide by the bottom numbers to get your answer!
25.3 moles H20
Step 6: Cancel out units that are both in the numerator and denominator and multiply.
x
1 mole H2O
18.0 grams H2O
=
1
455 grams H20
Next
Use balanced equation
1 Mole = 6.02 x 1023 particles
*Note: This simulation shows the immediate water displacement when a rainbow swirl is added to water, before it dissolves.
particlesB
1 Mole = 6.02 x 1023 particles
mass A(grams)
Use Molar Mass
particlesA
volume of gas A(liters)
molesB
Use balanced equation
1 Mole = 22.4 L of gas at STP
molesA
mass B(grams)
Use Molar Mass
volume of gas B(liters)
1 Mole = 22.4 L of gas at STP
Use balanced equation
20.0 moles H20 = ? molecules H20
Step 2: Draw a multiplication sign and another line.
20.0 moles H20
1
x
Step 1: Put what you start with over 1.
20.0 moles H20
1
Step 6: Cancel out units that are both in the numerator and denominator and multiply.
20.0 moles H2O
1
Step 3: Write the units you want to cancel out in the denominator of this new fraction.
x
mole H20
Step 4: Write the units of your next stop in the numerator.
x
mole H2O
molecules H20
20.0 moles H2O
1
Step 5: Use the corresponding conversion factor to write in the values.
x
1 mole H2O
6.02 x 1023 molecules H2O
20.0 moles H20
1
1 mole of H2O has 6.02 x 1023 molecules.
20.0 moles H20
x
1 mole H2O
6.02 x 1023 molecules H2O
=
1
1.20 x 1024 molecules H20
Step 6: Cancel out units that are in both the numerator and denominator. Then, multiply the top numbers and divide by the bottom numbers to get your answer!
Next
Use balanced equation
1 Mole = 6.02 x 1023 particles
*Note: This simulation shows the immediate water displacement when a rainbow swirl is added to water, before it dissolves.
particlesB
1 Mole = 6.02 x 1023 particles
mass A(grams)
Use Molar Mass
particlesA
volume of gas A(liters)
molesB
Use balanced equation
1 Mole = 22.4 L of gas at STP
molesA
mass B(grams)
Use Molar Mass
volume of gas B(liters)
1 Mole = 22.4 L of gas at STP
Use balanced equation
1.29 x 1024 Na atoms = ? moles of Na
Step 2: Draw a multiplication sign and another line.
20 moles H20
1
x
20 moles H2O
1
mole H20
mole H2O
20 moles H2O
1
x
20 moles H20
1
1.20 x 1024 molecules H20
Step 1: Put what you start with over 1.
1.29 x 1024 Na atoms
1
1.29 x 1024 Na atoms
Step 6: Cancel out units that are in both the numerator and denominator. Then, multiply the top numbers and divide by the bottom numbers to get your answer!
2.14 moles Na
Step 6: Cancel out units that are both in the numerator and denominator and multiply.
=
6.02 x 1023 Na atoms
1 mole Na
x
1
1.29 x 1024 Na atoms
Step 3: Write the units you want to cancel out in the denominator of this new fraction.
x
1
1.29 x 1024 Na atoms
Na atoms
Step 4: Write the units of your next stop in the numerator.
x
mole Na
1
1.29 x 1024 Na atoms
Na atoms
Step 5: Use the corresponding conversion factor to write in the values.
6.02 x 1023 Na atoms
1 mole of Na has 6.02 x 1023 atoms.
1 mole Na
x
1
1.29 x 1024 Na atoms
Next
Use balanced equation
1 Mole = 6.02 x 1023 particles
particlesB
1 Mole = 6.02 x 1023 particles
mass A(grams)
Use Molar Mass
particlesA
volume of gas A(liters)
molesB
Use balanced equation
1 Mole = 22.4 L of gas at STP
molesA
mass B(grams)
Use Molar Mass
volume of gas B(liters)
1 Mole = 22.4 L of gas at STP
Use balanced equation
5.60 Liters CH4 at STP= ? moles of CH4
Step 1: Put what you start with over 1.
5.60 L CH4
1
Step 2: Write a multiplication sign and another line.
1
x
5.60 L CH4
Step 3: Write the units you want to cancel out in the denominator of this new fraction.
x
L CH4
1
5.60 L CH4
Step 4: Write the units of your next stop in the numerator.
mole CH4
x
L CH4
1
5.60 L CH4
Step 5: Use the corresponding conversion factor to write in the values.
1 mole of any gas at STP has a volume of 22.4 L.
1 mole CH4
x
22.4 L CH4
1
5.60 L CH4
5.60 L CH4
0.250 moles CH4
Step 6: Cancel out units that are both in the numerator and denominator and multiply.
=
1 mole CH4
x
22.4 L CH4
1
Step 6: Cancel out units that are in both the numerator and denominator. Then, multiply the top numbers and divide by the bottom numbers to get your answer!
Next
*Note: This simulation shows the immediate water displacement when a rainbow swirl is added to water, before it dissolves.
12.0 moles of O2= ? Liters of O2 at STP
20 moles H20
20 moles H2O
1
mole H20
mole H2O
20 moles H2O
1
x
20 moles H20
1
Use balanced equation
1.20 x 1024 molecules H20
1 Mole = 6.02 x 1023 particles
1.29 x 1024 Na atoms
1
particlesB
1.29 x 1024 Na atoms
Na atoms
1 Mole = 6.02 x 1023 particles
20 moles H20
6.02 x 1023 Na atoms
x
mass A(grams)
Use Molar Mass
Step 1: Put what you start with over 1.
12.0 moles O2
particlesA
1
volume of gas A(liters)
Step 2: Draw a multiplication sign and another line.
molesB
x
Use balanced equation
1 Mole = 22.4 L of gas at STP
molesA
mass B(grams)
Use Molar Mass
volume of gas B(liters)
1 Mole = 22.4 L of gas at STP
Use balanced equation
L CH4
22.4 L CH4
12.0 moles O2
1
moles O2
12.0 moles O2
1
Step 3: Write the units you to cancel out in the denominator of this new fraction.
x
moles O2
Step 4: Write the units of your next stop in the numerator.
L O2
12.0 moles O2
1
x
mole O2
Step 5: Use the corresponding conversion factor to write in the values.
1 mole of any gas at STP has a volume of 22.4 L.
22.4 L O2
12.0 moles O2
1
x
1 mole O2
269 L O2
Step 6: Cancel out units that are both in the numerator and denominator and multiply.
=
22.4 L O2
12.0 moles O2
1
x
1 mole O2
Step 6: Cancel out units that are in both the numerator and denominator. Then, multiply the top numbers and divide by the bottom numbers to get your answer!
30.0 grams of H2O = ? Molecules of H2O
Next
Step 1: Put what you start with over 1.
30.0 g H2O
1
Step 4: Write the units of your next stop in the numerator.
mole H2O
30.0 g H2O
1
x
g H2O
Use balanced equation
1 Mole = 6.02 x 1023 particles
particlesB
1 Mole = 6.02 x 1023 particles
mass A(grams)
Use Molar Mass
particlesA
volume of gas A(liters)
molesB
Use balanced equation
1 Mole = 22.4 L of gas at STP
molesA
mass B(grams)
Use Molar Mass
volume of gas B(liters)
1 Mole = 22.4 L of gas at STP
Use balanced equation
30.0 g H2O
1
Step 3: Write the units you want to cancel out in the denominator of this new fraction.
x
g H2O
Step 2: Write a multiplication sign and another line.
x
30.0 g H2O
1
Step 5: Use the corresponding conversion factor to write in the values.
1 mole of H2O has a mass of 18.0 grams.
1 mole H2O
30.0 g H2O
1
x
18.0 g H2O
x
1 mole H2O
30.0 g H2O
1
x
18.0 g H2O
Repeat Step 2: Write a multiplication sign and another line.
Repeat Step 4: Write the units of your next stop in the numerator.
x
1 mole H2O
30.0 g H2O
1
x
18.0 g H2O
mole H2O
molecules H2O
Repeat Step 3: Write the units you want to cancel out in the denominator of this new fraction.
x
1 mole H2O
30.0 g H2O
1
x
18.0 g H2O
mole H2O
x
1 mole H2O
30.0 g H2O
1
x
18.0 g H2O
Repeat Step 5: Use the corresponding conversion factor to write in the values.
1 mole H2O
6.02 x 1023 molecules H2O
1 mole of H2O has 6.02 x 1023 molecules.
x
1 mole H2O
30.0 g H2O
1
x
18.0 g H2O
1 mole H2O
6.02 x 1023 molecules H2O
Step 6: Cancel out units that are both in the numerator and denominator and multiply.
1.00 x 1024 molecules H2O
=
Step 6: Cancel out units that are in both the numerator and denominator. Then, multiply the top numbers and divide by the bottom numbers to get your answer!
1 Mole = 22.4 L of gasat standard temperature & pressure
95.0 x 1023 K atoms = ? grams K
Next
mole H2O
95.0 x 1024 K atoms = ? grams K
Step 2: Draw a multiplication sign and another line.
x
1
Step 1: Put what you start with over 1.
95.0 x 1023 K atoms
1
95.0 x 1023 K atoms
Step 3: Write the units you want to cancel out in the denominator of this new fraction.
x
1
95.0 x 1023 K atoms
K atoms
Step 4: Write the units of your next stop in the numerator.
mole K
x
1
95.0 x 1023 K atoms
K atoms
Step 5: Use the corresponding conversion factor to write in the values.
1 mole K
x
1
95.0 x 1023 K atoms
1 mole of K contains 6.02 x 1023 K atoms
6.02 x 1023 K atoms
Repeat Step 2: Draw a multiplication sign and another line.
1 mole K
x
1
95.0 x 1023 K atoms
6.02 x 1023 K atoms
x
Repeat Step 3: Write the units you want to cancel out in the denominator of this new fraction.
1 mole K
x
1
95.0 x 1023 K atoms
6.02 x 1023 K atoms
x
mole K
1 mole K
x
1
95.0 x 1023 K atoms
6.02 x 1023 K atoms
x
mole K
Repeat Step 4: Write the units of your next stop in the numerator.
g K
Use balanced equation
1 Mole = 6.02 x 1023 particles
particlesB
1 Mole = 6.02 x 1023 particles
mass A(grams)
Use Molar Mass
particlesA
volume of gas A(liters)
molesB
Use balanced equation
1 Mole = 22.4 L of gas at STP
molesA
mass B(grams)
Use Molar Mass
volume of gas B(liters)
1 Mole = 22.4 L of gas at STP
Use balanced equation
1 mole of Khas a mass of 39.1 grams.
1 mole K
x
1
95.0 x 1023 K atoms
6.02 x 1023 K atoms
x
1 mole K
39.1 g K
Repeat Step 5: Use the corresponding conversion factor to write in the values.
1 mole K
x
1
95.0 x 1023 K atoms
6.02 x 1023 K atoms
x
1 mole K
39.1 g K
Step 6: Cancel out units that are both in the numerator and denominator and multiply.
617 g K
=
Step 6: Cancel out units that are both in the numerator and denominator. Then, multiply the top numbers and divide the bottom numbers to get your answer!
36.0 x 1023 CO2 molecules = ? Liters CO2 at STP
Next
Step 1: Put what you start with over 1.
36.0 x 1023 CO2 molecules
1
Step 2: Draw a multiplication sign and another line.
x
1
36.0 x 1023 CO2 molecules
Step 3: Write the units you want to cancel out in the denominator of this new fraction.
x
1
36.0 x 1023 CO2 molecules
CO2 molecules
Step 4: Write the units of your next stop in the numerator.
mole CO2
x
1
36.0 x 1023 CO2 molecules
CO2 molecules
Use balanced equation
1 Mole = 6.02 x 1023 particles
particlesB
1 Mole = 6.02 x 1023 particles
mass A(grams)
Use Molar Mass
particlesA
volume of gas A(liters)
molesB
Use balanced equation
1 Mole = 22.4 L of gas at STP
molesA
mass B(grams)
Use Molar Mass
volume of gas B(liters)
1 Mole = 22.4 L of gas at STP
Use balanced equation
Step 5: Use the corresponding conversion factor to write in the values.
1 mole of CO2 contains 6.02 x 1023 molecules.
1 mole CO2
x
1
36.0 x 1023 CO2 molecules
6.02 x 1023 CO2 molecules
Repeat Step 2: Draw a multiplication sign and another line.
x
1 mole CO2
x
1
36.0 x 1023 CO2 molecules
6.02 x 1023 CO2 molecules
Repeat Step 3: Write the units you want to cancel out in the denominator of this new fraction.
x
mole CO2
1 mole CO2
x
1
36.0 x 1023 CO2 molecules
6.02 x 1023 CO2 molecules
x
mole CO2
Repeat Step 4: Write the units of your next stop in the numerator.
L CO2
1 mole CO2
x
1
36.0 x 1023 CO2 molecules
6.02 x 1023 CO2 molecules
x
1 mole CO2
22.4 L CO2
1 mole CO2
x
1
34.0 x 1023 CO2 molecules
6.02 x 1023 CO2 molecules
1 mole of any gas at STP has a volume of 22.4 L.
Repeat Step 5: Use the corresponding conversion factor to write in the values.
Step 6: Cancel out units that are both in the numerator and denominator and multiply.
x
1 mole CO2
22.4 L CO2
1 mole CO2
x
1
34.0 x 1023 CO2 molecules
6.02 x 1023 CO2 molecules
134 L CO2
=
Step 6: Cancel out units that are in both the numerator and denominator. Then, multiply the top numbers and divide by the bottom numbers to get your answer!
7.50 Liters of CH4 at STP = ? Molecules CH4
Next
Use balanced equation
1 Mole = 6.02 x 1023 particles
particlesB
1 Mole = 6.02 x 1023 particles
mass A(grams)
Use Molar Mass
particlesA
volume of gas A(liters)
molesB
Use balanced equation
1 Mole = 22.4 L of gas at STP
molesA
mass B(grams)
Use Molar Mass
volume of gas B(liters)
1 Mole = 22.4 L of gas at STP
Use balanced equation
Step 1: Put what you start with over 1.
7.50 L CH4
1
Step 2: Draw a multiplication sign and another line.
x
7.50 L CH4
1
Step 3: Write the units you want to cancel out in the denominator of this new fraction.
x
1
36.0 x 1023 CO2 molecules
L CH4
7.50 L CH4
1
Step 4: Write the units of your next stop in the numerator.
mole CH4
CO2 molecules
x
L CH4
7.50 L CH4
1
1 mole CO2
x
1
36.0 x 1023 CO2 molecules
6.02 x 1023 CO2 molecules
Step 5: Use the corresponding conversion factor to write in the values.
1 mole of any gas at STP has a volume of 22.4 L.
1 mole CH4
x
22.4 L CH4
7.50 L CH4
1
Repeat Step 2: Draw a multiplication sign and another line.
x
1 mole CH4
x
22.4 L CH4
7.50 L CH4
1
Repeat Step 3: Write the units you want to cancel out in the denominator of this new fraction.
mole CH4
x
1 mole CH4
x
22.4 L CH4
7.50 L CH4
1
mole CO2
Repeat Step 4: Write the units of your next stop in the numerator.
molecules CH4
mole CH4
x
1 mole CH4
x
22.4 L CH4
7.50 L CH4
1
22.4 L CO2
There are 6.02 x 1023 molecules in a mole of CH4.
Repeat Step 5: Use the corresponding conversion factor to write in the values.
6.02 x 1023 molecules CH4
1 mole CH4
x
1 mole CH4
x
22.4 L CH4
7.50 L CH4
1
1 mole CO2
Step 6: Cancel out units that are both in the numerator and denominator and multiply.
2.02 x 1023 molecules CH4
=
6.02 x 1023 molecules CH4
1 mole CH4
x
1 mole CH4
x
22.4 L CH4
7.50 L CH4
1
Step 6: Cancel out units that are in both the numerator and denominator. Then, multiply the top numbers and divide by the bottom numbers to get your answer!
61.5 grams CO at STP = ? Liters CO
Next
Step 1: Put what you start with over 1.
61.5 g CO
1
Step 2: Draw a multiplication sign and another line.
x
61.5 g CO
1
Step 3: Write the units you want to cancel out in the denominator of this new fraction.
x
g CO
61.5 g CO
1
Step 4: Write the units of your next stop in the numerator.
mole CO
x
g CO
61.5 g CO
1
Step 5: Use the corresponding conversion factor to write in the values.
1 mole of CO has a mass of 28.0 grams.
1 mole CO
x
28.0 g CO
61.5 g CO
1
Repeat Step 2: Draw a multiplication sign and another line.
x
1 mole CO
x
28.0 g CO
61.5 g CO
1
Repeat Step 3: Write the units you want to cancel out in the denominator of this new fraction.
mole CO
x
1 mole CO
x
Use balanced equation
1 Mole = 6.02 x 1023 particles
particlesB
1 Mole = 6.02 x 1023 particles
mass A(grams)
Use Molar Mass
particlesA
volume of gas A(liters)
28.0 g CO
molesB
Use balanced equation
1 Mole = 22.4 L of gas at STP
molesA
mass B(grams)
Use Molar Mass
volume of gas B(liters)
61.5 g CO
1 Mole = 22.4 L of gas at STP
Use balanced equation
1
Repeat Step 4: Write the units of your next stop in the numerator.
L CO
mole CO
x
1 mole CO
x
28.0 g CO
61.5 g CO
1
One mole of any gas at STP has a volume of 22.4 L.
Repeat Step 5: Use the corresponding conversion factor to write in the values.
x
1 mole CO
x
28.0 g CO
61.5 g CO
1
22.4 L CO
1 mole CO
Step 6: Cancel out units that are both in the numerator and denominator and multiply.
61.5 g CO
1 mole CO
28.0 g CO
49.2 L CO
x
x
1
22.4 L CO
1 mole CO
=
Step 6: Cancel out units that are in both the numerator and denominator. Then, multiply the top numbers and divide by the bottom numbers to get your answer!
Use balanced equation
61.5 g CO
28.0 g N2
44.8 Liters N2 at STP = ? grams N2
Next
Step 1: Put what you start with over 1.
44.8 L N2
1
Step 2: Draw a multiplication sign and another line.
x
61.5 g CO
1
Step 3: Write the units you want to cancel out in the denominator of this new fraction.
x
g CO
61.5 g CO
1
Step 4: Write the units of your next stop in the numerator.
mole CO
x
g CO
61.5 g CO
1
Step 5: Use the corresponding conversion factor to write in the values.
1 mole of N2 has a mass of 28.0 grams.
1 mole N2
x
28.0 g CO
61.5 g CO
1
Repeat Step 2: Draw a multiplication sign and another line.
x
1 mole N2
x
28.0 g CO
61.5 g CO
1
Repeat Step 3: Write the units you want to cancel out in the denominator of this new fraction.
mole N2
x
1 mole N2
x
28.0 g CO
61.5 g CO
1
Repeat Step 4: Write the units of your next stop in the numerator.
g N2
mole N2
x
1 mole N2
x
28.0 g CO
61.5 g CO
1
One mole of any gas at STP has a volume of 22.4 L.
Repeat Step 5: Use the corresponding conversion factor to write in the values.
x
1 mole N2
x
28.0 g CO
61.5 g CO
1
22.4 L CO
1 mole N2
1 mole CO
44.8 L N2
44.8 L N2
44.8 L N2
44.8 L N2
44.8 L N2
44.8 L N2
44.8 L N2
44.8 L N2
22.4 L N2
22.4 L N2
22.4 L N2
22.4 L N2
L N2
L N2
1 Mole = 6.02 x 1023 particles
particlesB
1 Mole = 6.02 x 1023 particles
mass A(grams)
Use Molar Mass
particlesA
volume of gas A(liters)
molesB
Use balanced equation
1 Mole = 22.4 L of gas at STP
molesA
mass B(grams)
Use Molar Mass
volume of gas B(liters)
1 Mole = 22.4 L of gas at STP
Use balanced equation
1 mole N2
22.4 L N2
44.8 L N2
1 mole N2
Step 6: Cancel out units that are both in the numerator and denominator and multiply.
56.0 g N2
x
x
1
=
28.0 g N2
Step 6: Cancel out units that are in both the numerator and denominator. Then, multiply the top numbers and divide by the bottom numbers to get your answer!
Use balanced equation
61.5 g CO
How many moles of O2 are produced when 10.6 moles of KClO3 decompose?
Next
61.5 g CO
g CO
61.5 g CO
g CO
61.5 g CO
28.0 g CO
61.5 g CO
28.0 g CO
22.4 L CO
1 mole CO
1 Mole = 6.02 x 1023 particles
particlesB
1 Mole = 6.02 x 1023 particles
mass A(grams)
Use Molar Mass
particlesA
volume of gas A(liters)
molesB
Use balanced equation
1 Mole = 22.4 L of gas at STP
molesA
mass B(grams)
Use Molar Mass
volume of gas B(liters)
1 Mole = 22.4 L of gas at STP
Use balanced equation
2KClO3 → 2KCl + 3O2
L N2
28.0 g N2
1 mole of N2 has a mass of 28.0 grams.
Repeat Step 5: Use the corresponding conversion factor to write in the values.
x
1 mole N2
x
1
1 mole N2
44.8 L N2
22.4 L N2
Repeat Step 4: Write the units of your next stop in the numerator.
g N2
mole N2
x
1 mole N2
x
1
44.8 L N2
22.4 L N2
Repeat Step 3: Write the units you want to cancel out in the denominator of this new fraction.
mole N2
x
1 mole N2
x
28.0 g CO
1
44.8 L N2
Repeat Step 2: Draw a multiplication sign and another line.
x
1 mole N2
x
1
44.8 L N2
22.4 L N2
Step 1: Put what you start with over 1.
10.6 moles KClO3
1
Step 3: Write the units you want to cancel out in the denominator of this new fraction.
x
10.6 moles KClO3
1
moles KClO3
= 15.9 moles O2
Step 6: Cancel out units that are in both the numerator and denominator. Then, multiply the top numbers and divide by the bottom numbers to get your answer!
x
10.6 moles KClO3
1
2 moles KClO3
3 moles O2
Step 5: Use the corresponding conversion factor to write in the values.
x
10.6 moles KClO3
1
2 moles KClO3
3 moles O2
The balanced reaction shows the ratio to O2 to KClO3 is 3:2.
2KClO3 → 2KCl + 3O2
Step 4: Write the units of your next stop in the numerator.
x
10.6 moles KClO3
1
moles KClO3
moles O2
Step 2: Draw a multiplication sign and another line.
x
10.6 moles KClO3
1
Use balanced equation
61.5 g CO
How many liters of O2 are needed for the complete combustion of 11.5 liters of C3H8?
Next
61.5 g CO
g CO
61.5 g CO
g CO
61.5 g CO
28.0 g CO
61.5 g CO
28.0 g CO
22.4 L CO
1 mole CO
1 Mole = 6.02 x 1023 particles
particlesB
1 Mole = 6.02 x 1023 particles
mass A(grams)
Use Molar Mass
particlesA
volume of gas A(liters)
molesB
Use balanced equation
1 Mole = 22.4 L of gas at STP
molesA
mass B(grams)
Use Molar Mass
volume of gas B(liters)
1 Mole = 22.4 L of gas at STP
Use balanced equation
C3H8 + 5O2 → 3CO2 + 4H2O
L N2
28.0 g N2
1 mole of N2 has a mass of 28.0 grams.
Repeat Step 5: Use the corresponding conversion factor to write in the values.
x
1 mole N2
x
1
1 mole N2
44.8 L N2
22.4 L N2
Repeat Step 4: Write the units of your next stop in the numerator.
g N2
mole N2
x
1 mole N2
x
1
44.8 L N2
22.4 L N2
Repeat Step 3: Write the units you want to cancel out in the denominator of this new fraction.
mole N2
x
1 mole N2
x
28.0 g CO
1
44.8 L N2
Repeat Step 2: Draw a multiplication sign and another line.
x
1 mole N2
x
1
44.8 L N2
22.4 L N2
Step 1: Put what you start with over 1.
11.5 L C3H8
1
Step 2: Draw a multiplication sign and another line.
x
10.6 moles KClO3
1
Step 3: Write the units you want to cancel out in the denominator of this new fraction.
x
10.6 moles KClO3
1
L C3H8
Step 4: Write the units of your next stop in the numerator.
2KClO3 → 2KCl + 3O2
11.5 L C3H8
1
11.5 L C3H8
1
x
L C3H8
11.5 L C3H8
1
L O2
2000 SOUNDINGS CRESCENT COURT, SUFFOLK, VA 23435
Step 5: Use the corresponding conversion factor to write in the values.
The balanced reaction shows the ratio to O2 to C3H8 is 5:1.
1C3H8 + 5O2 → 3CO2 + 4H2O
x
1 L C3H8
11.5 L C3H8
1
5 L O2
= 57.5 L O2
Step 6: Cancel out units that are in both the numerator and denominator. Then, multiply the top numbers and divide by the bottom numbers to get your answer!
x
1 L C3H8
11.5 L C3H8
1
5 L O2
Use balanced equation
61.5 g CO
How many molecules of NH3 are produced when 45.8 x 1023 molecules of H2 completely react with N2?
Next
61.5 g CO
g CO
61.5 g CO
g CO
61.5 g CO
28.0 g CO
61.5 g CO
28.0 g CO
22.4 L CO
1 mole CO
1 Mole = 6.02 x 1023 particles
particlesB
1 Mole = 6.02 x 1023 particles
mass A(grams)
Use Molar Mass
particlesA
volume of gas A(liters)
molesB
Use balanced equation
1 Mole = 22.4 L of gas at STP
molesA
mass B(grams)
Use Molar Mass
volume of gas B(liters)
1 Mole = 22.4 L of gas at STP
Use balanced equation
3H2 + N2 → 2NH3
L N2
28.0 g N2
1 mole of N2 has a mass of 28.0 grams.
Repeat Step 5: Use the corresponding conversion factor to write in the values.
x
1 mole N2
x
1
1 mole N2
44.8 L N2
22.4 L N2
Repeat Step 4: Write the units of your next stop in the numerator.
g N2
mole N2
x
1 mole N2
x
1
44.8 L N2
22.4 L N2
Repeat Step 3: Write the units you want to cancel out in the denominator of this new fraction.
mole N2
x
1 mole N2
x
28.0 g CO
1
44.8 L N2
Repeat Step 2: Draw a multiplication sign and another line.
x
1 mole N2
x
1
44.8 L N2
22.4 L N2
10.6 moles KClO3
1
10.6 moles KClO3
1
2KClO3 → 2KCl + 3O2
2000 SOUNDINGS CRESCENT COURT, SUFFOLK, VA 23435
Step 1: Put what you start with over 1.
1
Step 2: Draw a multiplication sign and another line.
x
11.5 L C3H8
1
Step 3: Write the units you want to cancel out in the denominator of this new fraction.
x
molecules H2
Step 4: Write the units of your next stop in the numerator.
1
molecules NH3
1C3H8 + 5O2 → 3CO2 + 4H2O
45.8 x 1024 molecules H2
1
45.8 x 1024 molecules H2
1
45.8 x 1024 molecules H2
x
molecules H2
1
45.8 x 1024 molecules H2
= 30.5 x 1024 molecules NH3
Step 6: Cancel out units that are in both the numerator and denominator. Then, multiply the top numbers and divide by the bottom numbers to get your answer!
2 molecules NH3
x
3 molecules H2
1
45.8 x 1024 molecules H2
Step 5: Use the corresponding conversion factor to write in the values.
The balanced reaction shows the ratio to NH3 to H2 is 2:3.
2 molecules NH3
x
3 molecules H2
1
45.8 x 1024 molecules H2
3H2 + N2 → 2NH3
mole O2
x
2 mole KClO3
3 mole O2
Use balanced equation
44.8 Liters N2 at STP = ? grams N2
Next
61.5 g CO
g CO
61.5 g CO
g CO
61.5 g CO
28.0 g CO
61.5 g CO
28.0 g CO
61.5 g CO
61.5 g CO
28.0 g CO
61.5 g CO
28.0 g CO
61.5 g CO
1 Mole = 6.02 x 1023 particles
particlesB
1 Mole = 6.02 x 1023 particles
mass A(grams)
Use Molar Mass
particlesA
volume of gas A(liters)
molesB
Use balanced equation
1 Mole = 22.4 L of gas at STP
molesA
mass B(grams)
Use Molar Mass
volume of gas B(liters)
1 Mole = 22.4 L of gas at STP
Use balanced equation
Step 6: Cancel out units that are both in the numerator and denominator and multiply.
Step 1: Put what you start with over 1.
96.2 g KClO3
1
Step 2: Draw a multiplication sign and another line.
x
Step 3: Write the units you want to cancel out in the denominator of this new fraction.
x
g KClO3
Step 4: Write the units of your next stop in the numerator.
mole KClO3
= 37.7 g O2
Step 5: Use the corresponding conversion factor to write in the values.
One mole of KClO3 has a mass of 122.6 grams.
Repeat Step 2: Draw a multiplication sign and another line.
x
Repeat Step 3: Write the units you want to cancel out in the denominator of this new fraction.
mole N2
x
Step 6: Cancel out units that are in both the numerator and denominator. Then, multiply the top numbers and divide by the bottom numbers to get your answer!
Repeat Step 4: Write the units of your next stop in the numerator.
g N2
mole KClO3
1 mole O2
1 mole N2
22.4 L N2
44.8 L N2
x
1 mole N2
2 mole KClO3
3 mole O2
x
1 mole KClO3
x
122.6 g KClO3
96.2 g KClO3
1
32.0 g O2
x
x
1
=
28.0 g N2
2KClO3 → 2KCl + 3O2
How many grams of O2 are produced when 96.2 grams of KClO3 decompose?
96.2 g KClO3
1
96.2 g KClO3
1
x
g KClO3
96.2 g KClO3
1
1 mole KClO3
x
122.6 g KClO3
96.2 g KClO3
1
1 mole KClO3
x
122.6 g KClO3
96.2 g KClO3
1
1 mole KClO3
x
122.6 g KClO3
96.2 g KClO3
1
1 mole KClO3
x
122.6 g KClO3
96.2 g KClO3
1
x
mole O2
mole KClO3
Repeat Step 3: Write the units you want to cancel out in the denominator of this new fraction.
x
1 mole KClO3
x
122.6 g KClO3
96.2 g KClO3
1
mole KClO3
Repeat Step 4: Write the units of your next stop in the numerator.
mole O2
x
2 mole KClO3
3 mole O2
x
1 mole KClO3
x
122.6 g KClO3
96.2 g KClO3
1
g O2
The balanced reaction shows the ratio of O2 to KClO3 is 3:2.
Repeat Step 5: Use the corresponding conversion factor to write in the values.
2 mole KClO3
1 mole KClO3
x
122.6 g KClO3
96.2 g KClO3
1
x
3 mole O2
2KClO3 → 2KCl + 3O2
Repeat Step 2: Draw a multiplication sign and another line.
x
2 mole KClO3
1 mole KClO3
x
122.6 g KClO3
96.2 g KClO3
1
x
3 mole O2
Repeat Step 5: Use the corresponding conversion factor to write in the values.
1 mole O2
x
2 mole KClO3
3 mole O2
x
1 mole KClO3
x
122.6 g KClO3
96.2 g KClO3
1
32.0 g O2
1 mole of O2has a mas of 32.0 grams.
2KClO3 → 2KCl + 3O2
mole O2
x
2 mole KClO3
3 mole O2
Use balanced equation
Next
1 Mole = 6.02 x 1023 particles
particlesB
1 Mole = 6.02 x 1023 particles
mass A(grams)
Use Molar Mass
particlesA
volume of gas A(liters)
molesB
Use balanced equation
1 Mole = 22.4 L of gas at STP
molesA
mass B(grams)
Use Molar Mass
volume of gas B(liters)
1 Mole = 22.4 L of gas at STP
Use balanced equation
Step 1: Put what you start with over 1.
44.8 g KClO3
1
Step 2: Draw a multiplication sign and another line.
x
Step 3: Write the units you want to cancel out in the denominator of this new fraction.
x
g KClO3
Step 4: Write the units of your next stop in the numerator.
mole KClO3
Step 5: Use the corresponding conversion factor to write in the values.
One mole of KClO3 has a mass of 122.6 grams.
Repeat Step 2: Draw a multiplication sign and another line.
x
Repeat Step 3: Write the units you want to cancel out in the denominator of this new fraction.
x
Repeat Step 4: Write the units of your next stop in the numerator.
mole KClO3
2KClO3 → 2KCl + 3O2
How many liters of O2 (at standard temperatuure and pressure) are produced when 44.8 grams of KClO3 decompose?
44.8 g KClO3
1
44.8 g KClO3
1
x
g KClO3
44.8 g KClO3
1
1 mole KClO3
x
122.6 g KClO3
44.8 g KClO3
1
1 mole KClO3
x
122.6 g KClO3
44.8 g KClO3
1
1 mole KClO3
x
122.6 g KClO3
44.8 g KClO3
1
1 mole KClO3
x
122.6 g KClO3
44.8 g KClO3
1
x
mole O2
mole KClO3
Repeat Step 3: Write the units you want to cancel out in the denominator of this new fraction.
x
1 mole KClO3
x
122.6 g KClO3
44.8 g KClO3
1
Repeat Step 4: Write the units of your next stop in the numerator.
mole O2
x
2 mole KClO3
3 mole O2
x
1 mole KClO3
x
122.6 g KClO3
44.8 g KClO3
1
L O2
The balanced reaction shows the ratio of O2 to KClO3 is 3:2.
Repeat Step 5: Use the corresponding conversion factor to write in the values.
2 mole KClO3
1 mole KClO3
x
122.6 g KClO3
44.8 g KClO3
1
x
3 mole O2
2KClO3 → 2KCl + 3O2
Repeat Step 2: Draw a multiplication sign and another line.
x
2 mole KClO3
1 mole KClO3
x
122.6 g KClO3
44.8 g KClO3
1
x
3 mole O2
Repeat Step 5: Use the corresponding conversion factor to write in the values.
1 mole O2
x
2 mole KClO3
3 mole O2
x
1 mole KClO3
x
122.6 g KClO3
44.8 g KClO3
1
22.4 L O2
1 mole of any gas at STP has a volume of 22.4 L.
= 12.3 L O2
Step 6: Cancel out units that are in both the numerator and denominator. Then, multiply the top numbers and divide by the bottom numbers to get your answer!
1 mole O2
x
2 mole KClO3
3 mole O2
x
1 mole KClO3
x
122.6 g KClO3
44.8 g KClO3
1
22.4 L O2
2H2 + O2 → 2H2O
2KClO3 → 2KCl + 3O2
Use balanced equation
Next
1 Mole = 6.02 x 1023 particles
particlesB
1 Mole = 6.02 x 1023 particles
mass A(grams)
Use Molar Mass
particlesA
volume of gas A(liters)
molesB
Use balanced equation
1 Mole = 22.4 L of gas at STP
molesA
mass B(grams)
Use Molar Mass
volume of gas B(liters)
1 Mole = 22.4 L of gas at STP
Use balanced equation
Step 1: Put what you start with over 1.
215 g KClO3
1
Step 2: Draw a multiplication sign and another line.
x
Step 3: Write the units you want to cancel out in the denominator of this new fraction.
x
g KClO3
Step 4: Write the units of your next stop in the numerator.
mole KClO3
Step 5: Use the corresponding conversion factor to write in the values.
One mole of KClO3 has a mass of 122.6 grams.
Repeat Step 2: Draw a multiplication sign and another line.
x
Repeat Step 3: Write the units you want to cancel out in the denominator of this new fraction.
x
Repeat Step 4: Write the units of your next stop in the numerator.
mole KClO3
2KClO3 → 2KCl + 3O2
How many molecules of O2 are produced when 215 grams of KClO3 decompose?
215 g KClO3
1
215 g KClO3
1
x
g KClO3
215 g KClO3
1
1 mole KClO3
x
122.6 g KClO3
215 g KClO3
1
1 mole KClO3
x
122.6 g KClO3
215 g KClO3
1
1 mole KClO3
x
122.6 g KClO3
215 g KClO3
1
1 mole KClO3
x
122.6 g KClO3
215 g KClO3
1
x
mole O2
mole KClO3
The balanced reaction shows the ratio of O2 to KClO3 is 3:2.
Repeat Step 5: Use the corresponding conversion factor to write in the values.
2 mole KClO3
1 mole KClO3
x
122.6 g KClO3
215 g KClO3
1
x
3 mole O2
2KClO3 → 2KCl + 3O2
Repeat Step 5: Use the corresponding conversion factor to write in the values.
1 mole O2
x
2 mole KClO3
3 mole O2
x
1 mole KClO3
x
122.6 g KClO3
215 g KClO3
1
6.02 x 1023 molecules O2
1 mole of O2 conntains 6.02 x 1023 molecules.
mole O2
x
2 mole KClO3
3 mole O2
Repeat Step 3: Write the units you want to cancel out in the denominator of this new fraction.
x
1 mole KClO3
x
122.6 g KClO3
215 g KClO3
1
Repeat Step 2: Draw a multiplication sign and another line.
x
2 mole KClO3
1 mole KClO3
x
122.6 g KClO3
215 g KClO3
1
x
3 mole O2
Repeat Step 4: Write the units of your next stop in the numerator.
mole O2
x
2 mole KClO3
3 mole O2
x
1 mole KClO3
x
122.6 g KClO3
215 g KClO3
1
molecules O2
= 1.58 x 1024 molecules O2
Step 6: Cancel out units that are in both the numerator and denominator. Then, multiply the top numbers and divide by the bottom numbers to get your answer!
1 mole O2
x
2 mole KClO3
3 mole O2
x
1 mole KClO3
x
122.6 g KClO3
215 g KClO3
1
22.4 L O2
6.02 x 1023 molecules O2
Use balanced equation
Next
1 Mole = 6.02 x 1023 particles
particlesB
1 Mole = 6.02 x 1023 particles
mass A(grams)
Use Molar Mass
particlesA
volume of gas A(liters)
molesB
Use balanced equation
1 Mole = 22.4 L of gas at STP
molesA
mass B(grams)
Use Molar Mass
volume of gas B(liters)
1 Mole = 22.4 L of gas at STP
Use balanced equation
Step 1: Put what you start with over 1.
625 g KClO3
1
Step 2: Draw a multiplication sign and another line.
x
Step 3: Write the units you want to cancel out in the denominator of this new fraction.
x
g KClO3
Step 4: Write the units of your next stop in the numerator.
mole KClO3
Step 5: Use the corresponding conversion factor to write in the values.
One mole of KClO3 has a mass of 122.6 grams.
Repeat Step 2: Draw a multiplication sign and another line.
x
Repeat Step 3: Write the units you want to cancel out in the denominator of this new fraction.
x
Repeat Step 4: Write the units of your next stop in the numerator.
mole KClO3
2KClO3 → 2KCl + 3O2
How many moles of O2 are produced when 625 grams of KClO3 decompose?
625 g KClO3
1
625 g KClO3
1
x
g KClO3
625 g KClO3
1
1 mole KClO3
x
122.6 g KClO3
625 g KClO3
1
1 mole KClO3
x
122.6 g KClO3
625 g KClO3
1
1 mole KClO3
x
122.6 g KClO3
625 g KClO3
1
1 mole KClO3
x
122.6 g KClO3
625 g KClO3
1
x
mole O2
mole KClO3
The balanced reaction shows the ratio of O2 to KClO3 is 3:2.
Repeat Step 5: Use the corresponding conversion factor to write in the values.
2 mole KClO3
1 mole KClO3
x
122.6 g KClO3
625 g KClO3
1
x
3 mole O2
2KClO3 → 2KCl + 3O2
22.4 L O2
= 7.65 moles O2
Step 6: Cancel out units that are in both the numerator and denominator. Then, multiply the top numbers and divide by the bottom numbers to get your answer!
2 mole KClO3
3 mole O2
x
1 mole KClO3
x
122.6 g KClO3
625 g KClO3
1
Next
Step 1: Put what you start with over 1.
128 L H2
1
Step 2: Draw a multiplication sign and another line.
x
Step 3: Write the units you want to cancel out in the denominator of this new fraction.
x
L H2
Step 4: Write the units of your next stop in the numerator.
mole H2
Step 5: Use the corresponding conversion factor to write in the values.
One mole of any gas at STP has a volume of 22.4 L
Repeat Step 2: Draw a multiplication sign and another line.
x
Repeat Step 3: Write the units you want to cancel out in the denominator of this new fraction.
x
Repeat Step 4: Write the units of your next stop in the numerator.
mole H2
2H2 + O2 → 2H2O
How many moles of O2 are needed for the combustion of 128 L of H2 at standard temperature and pressure?
128 L H2
1
128 L H2
1
x
L H2
128 L H2
1
1 mole H2
x
22.4 L H2
128 L H2
1
1 mole H2
x
22.4 L H2
128 L H2
1
1 mole H2
x
22.4 L H2
128 L H2
1
1 mole H2
x
22.4 L H2
128 H2
1
x
mole O2
mole H2
The balanced reaction shows the ratio of O2 to H2 is 1:2.
Repeat Step 5: Use the corresponding conversion factor to write in the values.
2 mole H2
1 mole H2
x
22.4 L H2
128 H2
1
x
1 mole O2
2H2 + 1O2 → 2H2O
22.4 L O2
Use balanced equation
1 Mole = 6.02 x 1023 particles
particlesB
1 Mole = 6.02 x 1023 particles
mass A(grams)
Use Molar Mass
particlesA
volume of gas A(liters)
molesB
Use balanced equation
1 Mole = 22.4 L of gas at STP
molesA
mass B(grams)
Use Molar Mass
volume of gas B(liters)
1 Mole = 22.4 L of gas at STP
Use balanced equation
= 2.86 moles O2
Step 6: Cancel out units that are in both the numerator and denominator. Then, multiply the top numbers and divide by the bottom numbers to get your answer!
2 mole H2
1 mole O2
x
1 mole H2
x
22.4 L H2
128 L H2
1
Use balanced equation
Next
1 Mole = 6.02 x 1023 particles
particlesB
1 Mole = 6.02 x 1023 particles
mass A(grams)
Use Molar Mass
particlesA
volume of gas A(liters)
molesB
Use balanced equation
1 Mole = 22.4 L of gas at STP
molesA
mass B(grams)
Use Molar Mass
volume of gas B(liters)
1 Mole = 22.4 L of gas at STP
Use balanced equation
Step 1: Put what you start with over 1.
60.0 L O2
1
Step 2: Draw a multiplication sign and another line.
x
Step 3: Write the units you want to cancel out in the denominator of this new fraction.
x
L O2
Step 4: Write the units of your next stop in the numerator.
mole O2
Step 5: Use the corresponding conversion factor to write in the values.
The volume of 1 mole ofany gas at STP is 22.4 L.
Repeat Step 2: Draw a multiplication sign and another line.
x
Repeat Step 3: Write the units you want to cancel out in the denominator of this new fraction.
x
Repeat Step 4: Write the units of your next stop in the numerator.
mole O2
How many grams of H2O are produced when 60.0 L of O2 at standard temperature and pressure completely react with H2?
60.0 L O2
1
60.0 L O2
1
x
L O2
60.0 L O2
1
1 mole O2
x
22.4 L O2
60.0 L O2
1
1 mole O2
x
22.4 L O2
60.0 L O2
1
1 mole O2
x
22.4 L O2
60.0 L O2
1
1 mole O2
x
22.4 L O2
60.0 L O2
1
x
mole H2O
mole O2
The balanced reaction shows the ratio of H2O to O2 is 2:1.
Repeat Step 5: Use the corresponding conversion factor to write in the values.
1 mole O2
1 mole O2
x
22.4 L O2
60.0 L O2
1
x
2 moles H2O
2H2 + 1O2 → 2H2O
2H2 + O2 → 2H2O
Repeat Step 2: Draw a multiplication sign and another line.
x
1 mole O2
1 mole O2
x
22.4 L O2
60.0 L O2
1
x
2 moles H20
mole H2O
x
1 mole O2
2 moles H20
Repeat Step 3: Write the units you want to cancel out in the denominator of this new fraction.
x
1 mole O2
x
22.4 L O2
60.0 L O2
1
Repeat Step 4: Write the units of your next stop in the numerator.
mole H2O
x
1 mole O2
2 moles H2O
x
1 mole O2
x
22.4 L O2
60.0 L O2
1
g H2O
Repeat Step 5: Use the corresponding conversion factor to write in the values.
1 mole H2O
x
1 mole O2
2 moles H2O
x
1 mole O2
x
22.4 L O2
60.0 L O2
1
18.0 g H2O
1 mole of H2O has a mass of 18.0 grams.
= 96.4 g H2O
Step 6: Cancel out units that are in both the numerator and denominator. Then, multiply the top numbers and divide by the bottom numbers to get your answer!
1 mole H2O
x
1 mole O2
2 mole H2O
x
1 mole O2
x
22.4 L O2
60.0 L O2
1
18.0 g H2O
Use balanced equation
Next
1 Mole = 6.02 x 1023 particles
particlesB
1 Mole = 6.02 x 1023 particles
mass A(grams)
Use Molar Mass
particlesA
volume of gas A(liters)
molesB
Use balanced equation
1 Mole = 22.4 L of gas at STP
molesA
mass B(grams)
Use Molar Mass
volume of gas B(liters)
1 Mole = 22.4 L of gas at STP
Use balanced equation
Step 1: Put what you start with over 1.
30.0 L O2
1
Step 2: Draw a multiplication sign and another line.
x
Step 3: Write the units you want to cancel out in the denominator of this new fraction.
x
L O2
Step 4: Write the units of your next stop in the numerator.
mole O2
Step 5: Use the corresponding conversion factor to write in the values.
The volume of 1 mole of any gas at STP is 22.4 L.
Repeat Step 2: Draw a multiplication sign and another line.
x
Repeat Step 3: Write the units you want to cancel out in the denominator of this new fraction.
x
Repeat Step 4: Write the units of your next stop in the numerator.
mole O2
How many molecules of H2O are produced when 30.0 L of O2 at standard temperature and pressure completely react with H2?
30.0 L O2
1
30.0 L O2
1
x
L O2
30.0 L O2
1
1 mole O2
x
22.4 L O2
30.0 L O2
1
1 mole O2
x
22.4 L O2
30.0 L O2
1
1 mole O2
x
22.4 L O2
30.0 L O2
1
1 mole O2
x
22.4 L O2
30.0 L O2
1
x
mole H2O
mole O2
The balanced reaction shows the ratio of H2O to O2 is 2:1.
Repeat Step 5: Use the corresponding conversion factor to write in the values.
1 mole O2
1 mole O2
x
22.4 L O2
30.0 L O2
1
x
2 moles H2O
2H2 + 1O2 → 2H2O
2H2 + O2 → 2H2O
Repeat Step 2: Draw a multiplication sign and another line.
x
1 mole O2
1 mole O2
x
22.4 L O2
30.0 L O2
1
x
2 moles H20
= 1.61 x 1024 molecules H2O
Step 6: Cancel out units that are in both the numerator and denominator. Then, multiply the top numbers and divide by the bottom numbers to get your answer!
1 mole H2O
x
1 mole O2
2 mole H2O
x
1 mole O2
x
22.4 L O2
30.0 L O2
1
6.02 x 1023 molecules H2O
Repeat Step 5: Use the corresponding conversion factor to write in the values.
x
1 mole O2
2 moles H2O
x
1 mole O2
x
22.4 L O2
30.0 L O2
1
1 mole of H2O contains6.02 x 1023 molecules.
1 mole H2O
6.02 x 1023 molecules H2O
Repeat Step 4: Write the units of your next stop in the numerator.
x
1 mole O2
2 moles H2O
x
1 mole O2
x
22.4 L O2
30.0 L O2
1
mole H2O
molecules H2O
mole H2O
x
1 mole O2
2 moles H20
Repeat Step 3: Write the units you want to cancel out in the denominator of this new fraction.
x
1 mole O2
x
22.4 L O2
30.0 L O2
1
Next
2C8H18 + 25O2 → 16CO2 + 18H2O
How many moles of CO2 are produced as a result of the combustion of 5.80 x 1024 molecules of C8H18?
22.4 L O2
Use balanced equation
1 Mole = 6.02 x 1023 particles
particlesB
1 Mole = 6.02 x 1023 particles
mass A(grams)
Use Molar Mass
particlesA
volume of gas A(liters)
molesB
Use balanced equation
1 Mole = 22.4 L of gas at STP
molesA
mass B(grams)
Use Molar Mass
volume of gas B(liters)
1 Mole = 22.4 L of gas at STP
Use balanced equation
Step 1: Put what you start with over 1.
5.80 x 1024 molecules C8H18
1
Step 2: Draw a multiplication sign and another line.
x
5.80 x 1024 molecules C8H18
1
mole H2
Step 3: Write the units you want to cancel out in the denominator of this new fraction.
x
5.80 x 1024 molecules C8H18
1
molecules C8H18
Step 4: Write the units of your next stop in the numerator.
x
5.80 x 1024 molecules C8H18
1
molecules C8H18
mole C8H18
Step 5: Use the corresponding conversion factor to write in the values.
x
5.80 x 1024 molecules C8H18
6.02 x 1023 molecules C8H18
1 mole C8H18
1
One mole of C8H18 = 6.02 x 1023 molecules of C8H18.
Repeat Step 2: Draw a multiplication sign and another line.
x
x
5.80 x 1024 molecules C8H18
6.02 x 1023 molecules C8H18
1 mole C8H18
1
Repeat Step 3: Write the units you want to cancel out in the denominator of this new fraction.
moles C8H18
x
x
5.80 x 1024 molecules C8H18
6.02 x 1023 molecules C8H18
1 mole C8H18
1
Repeat Step 4: Write the units of your next stop in the numerator.
moles CO2
moles C8H18
x
x
5.80 x 1024 molecules C8H18
6.02 x 1023 molecules C8H18
1 mole C8H18
1
2H2 + 1O2 → 2H2O
Repeat Step 5: Use the corresponding conversion factor to write in the values.
16 moles CO2
2 moles C8H18
x
x
5.80 x 1024 molecules C8H18
6.02 x 1023 molecules C8H18
1 mole C8H18
1
2C8H18 + 25O2 → 16CO2 + 18H2O
The balanced reaction shows the ratio of CO2 to C8H18 is 16:2.
16 moles CO2
2 moles C8H18
x
x
5.80 x 1024 molecules C8H18
6.02 x 1023 molecules C8H18
1 mole C8H18
1
= 77.1 moles CO2
Step 6: Cancel out units that are in both the numerator and denominator. Then, multiply the top numbers and divide by the bottom numbers to get your answer!
2 mole H2
3 mole O2
x
1 mole H2
x
22.4 L H2
128 L H2
1
Next
2C8H18 + 25O2 → 16CO2 + 18H2O
How many grams of CO2 are produced as a result of the combustion of 2.60 x 1024 molecules of C8H18?
Use balanced equation
1 Mole = 6.02 x 1023 particles
particlesB
1 Mole = 6.02 x 1023 particles
mass A(grams)
Use Molar Mass
particlesA
volume of gas A(liters)
molesB
Use balanced equation
1 Mole = 22.4 L of gas at STP
molesA
mass B(grams)
Use Molar Mass
volume of gas B(liters)
1 Mole = 22.4 L of gas at STP
Use balanced equation
Repeat Step 5: Use the corresponding conversion factor to write in the values.
16 moles CO2
2 moles C8H18
x
x
2.60 x 1024 molecules C8H18
6.02 x 1023 molecules C8H18
1 mole C8H18
1
2C8H18 + 25O2 → 16CO2 + 18H2O
The balanced reaction shows the ratio of CO2 to C8H18 is 16:2.
Repeat Step 5: Use the corresponding conversion factor to write in the values.
1 mole of CO2 has a mass of 44.0 grams.
1 mole CO2
16 moles CO2
2 moles C8H18
x
x
2.60 x 1024 molecules C8H18
6.02 x 1023 molecules C8H18
1 mole C8H18
1
x
44.0 g CO2
Repeat Step 4: Write the units of your next stop in the numerator.
mole CO2
16 moles CO2
2 moles C8H18
x
x
2.60 x 1024 molecules C8H18
6.02 x 1023 molecules C8H18
1 mole C8H18
1
x
g CO2
Repeat Step 3: Write the units you want to cancel out in the denominator of this new fraction.
mole CO2
16 moles CO2
2 moles C8H18
x
x
2.60 x 1024 molecules C8H18
6.02 x 1023 molecules C8H18
1 mole C8H18
1
x
16 moles CO2
2 moles C8H18
x
x
2.60 x 1024 molecules C8H18
6.02 x 1023 molecules C8H18
1 mole C8H18
1
Repeat Step 2: Draw a multiplication sign and another line.
x
Repeat Step 4: Write the units of your next stop in the numerator.
moles CO2
moles C8H18
x
x
2.60 x 1024 molecules C8H18
6.02 x 1023 molecules C8H18
1 mole C8H18
1
Repeat Step 3: Write the units you want to cancel out in the denominator of this new fraction.
moles C8H18
x
x
2.60 x 1024 molecules C8H18
6.02 x 1023 molecules C8H18
1 mole C8H18
1
Repeat Step 2: Draw a multiplication sign and another line.
x
x
2.60 x 1024 molecules C8H18
6.02 x 1023 molecules C8H18
1 mole C8H18
1
Step 4: Write the units of your next stop in the numerator.
x
2.60 x 1024 molecules C8H18
1
molecules C8H18
mole C8H18
Step 3: Write the units you want to cancel out in the denominator of this new fraction.
x
2.60 x 1024 molecules C8H18
1
molecules C8H18
Step 2: Draw a multiplication sign and another line.
x
2.60 x 1024 molecules C8H18
1
Step 1: Put what you start with over 1.
2.60 x 1024 molecules C8H18
1
Step 5: Use the corresponding conversion factor to write in the values.
x
2.60 x 1024 molecules C8H18
6.02 x 1023 molecules C8H18
1 mole C8H18
1
One mole of C8H18 = 6.02 x 1023 molecules of C8H18.
1 mole CO2
16 moles CO2
2 moles C8H18
x
x
2.60 x 1024 molecules C8H18
6.02 x 1023 molecules C8H18
1 mole C8H18
1
x
44.0 g CO2
= 1520 g CO2
Step 6: Cancel out units that are in both the numerator and denominator. Then, multiply the top numbers and divide by the bottom numbers to get your answer!
Next
2C8H18 + 25O2 → 16CO2 + 18H2O
How many liters (at STP) of CO2 are produced as a result of the combustion of 7.60 x 1024 molecules of C8H18?
Use balanced equation
1 Mole = 6.02 x 1023 particles
particlesB
1 Mole = 6.02 x 1023 particles
mass A(grams)
Use Molar Mass
particlesA
volume of gas A(liters)
molesB
Use balanced equation
1 Mole = 22.4 L of gas at STP
molesA
mass B(grams)
Use Molar Mass
volume of gas B(liters)
1 Mole = 22.4 L of gas at STP
Use balanced equation
Repeat Step 5: Use the corresponding conversion factor to write in the values.
16 moles CO2
2 moles C8H18
x
x
7.60 x 1024 molecules C8H18
6.02 x 1023 molecules C8H18
1 mole C8H18
1
2C8H18 + 25O2 → 16CO2 + 18H2O
The balanced reaction shows the ratio of CO2 to C8H18 is 16:2.
Repeat Step 3: Write the units you want to cancel out in the denominator of this new fraction.
mole CO2
16 moles CO2
2 moles C8H18
x
x
7.60 x 1024 molecules C8H18
6.02 x 1023 molecules C8H18
1 mole C8H18
1
x
16 moles CO2
2 moles C8H18
x
x
7.60 x 1024 molecules C8H18
6.02 x 1023 molecules C8H18
1 mole C8H18
1
Repeat Step 2: Draw a multiplication sign and another line.
x
Repeat Step 4: Write the units of your next stop in the numerator.
moles CO2
moles C8H18
x
x
7.60 x 1024 molecules C8H18
6.02 x 1023 molecules C8H18
1 mole C8H18
1
Repeat Step 3: Write the units you want to cancel out in the denominator of this new fraction.
moles C8H18
x
x
7.60 x 1024 molecules C8H18
6.02 x 1023 molecules C8H18
1 mole C8H18
1
Repeat Step 2: Draw a multiplication sign and another line.
x
x
7.60 x 1024 molecules C8H18
6.02 x 1023 molecules C8H18
1 mole C8H18
1
Step 4: Write the units of your next stop in the numerator.
x
7.60 x 1024 molecules C8H18
1
molecules C8H18
mole C8H18
Step 3: Write the units you want to cancel out in the denominator of this new fraction.
x
7.60 x 1024 molecules C8H18
1
molecules C8H18
Step 2: Draw a multiplication sign and another line.
x
7.60 x 1024 molecules C8H18
1
Step 1: Put what you start with over 1.
7.60 x 1024 molecules C8H18
1
Step 5: Use the corresponding conversion factor to write in the values.
x
7.60 x 1024 molecules C8H18
6.02 x 1023 molecules C8H18
1 mole C8H18
1
One mole of C8H18 = 6.02 x 1023 molecules of C8H18.
Repeat Step 4: Write the units of your next stop in the numerator.
mole CO2
16 moles CO2
2 moles C8H18
x
x
7.60 x 1024 molecules C8H18
6.02 x 1023 molecules C8H18
1 mole C8H18
1
x
L CO2
Repeat Step 5: Use the corresponding conversion factor to write in the values.
1 mole of any gas at STP has a volume of 22.4 L.
1 mole CO2
16 moles CO2
2 moles C8H18
x
x
7.60 x 1024 molecules C8H18
6.02 x 1023 molecules C8H18
1 mole C8H18
1
x
22.4 L CO2
1 mole CO2
16 moles CO2
2 moles C8H18
x
x
7.60 x 1024 molecules C8H18
6.02 x 1023 molecules C8H18
1 mole C8H18
1
x
22.4 L CO2
= 2260 L CO2
Step 6: Cancel out units that are in both the numerator and denominator. Then, multiply the top numbers and divide by the bottom numbers to get your answer!
Use balanced equation
Next
1 Mole = 6.02 x 1023 particles
particlesB
1 Mole = 6.02 x 1023 particles
mass A(grams)
Use Molar Mass
particlesA
volume of gas A(liters)
molesB
Use balanced equation
1 Mole = 22.4 L of gas at STP
molesA
mass B(grams)
Use Molar Mass
volume of gas B(liters)
1 Mole = 22.4 L of gas at STP
Use balanced equation
How many liters of O2 (at standard temperatuure and pressure) are produced when 25.5 moles of KClO3 decompose?
2KClO3 → 2KCl + 3O2
Step 1: Put what you start with over 1.
25.5 moles KClO3
1
Step 2: Draw a multiplication sign and another line.
x
25.5 moles KClO3
1
Step 3: Write the units you want to cancel out in the denominator of this new fraction.
moles KClO3
x
25.5 moles KClO3
1
moles O2
One mole of KClO3 has a mass of 122.6 grams.
Step 4: Write the units of your next stop in the numerator.
moles KClO3
x
25.5 moles KClO3
1
Step 5: Use the corresponding conversion factor to write in the values.
3 moles O2
2 moles KClO3
x
25.5 moles KClO3
1
The balanced reaction shows the ratio of O2 to KClO3 is 3:2.
2KClO3 → 2KCl + 3O2
Repeat Step 2: Draw a multiplication sign and another line.
x
2 mole KClO3
1 mole KClO3
x
122.6 g KClO3
44.8 g KClO3
1
x
3 mole O2
3 moles O2
2 moles KClO3
x
25.5 moles KClO3
1
mole O2
Repeat Step 3: Write the units you want to cancel out in the denominator of this new fraction.
x
3 moles O2
2 moles KClO3
x
25.5 moles KClO3
1
Repeat Step 4: Write the units of your next stop in the numerator.
mole O2
x
L O2
3 moles O2
2 moles KClO3
x
25.5 moles KClO3
1
Repeat Step 5: Use the corresponding conversion factor to write in the values.
1 mole O2
x
22.4 L O2
The volume of 1 mole of any gas at STP is 22.4 L.
3 moles O2
2 moles KClO3
x
25.5 moles KClO3
1
= 857 L O2
Step 6: Cancel out units that are in both the numerator and denominator. Then, multiply the top numbers and divide by the bottom numbers to get your answer!
1 mole O2
x
22.4 L O2
3 moles O2
2 moles KClO3
x
25.5 moles KClO3
1
Use balanced equation
Next
1 Mole = 6.02 x 1023 particles
particlesB
1 Mole = 6.02 x 1023 particles
mass A(grams)
Use Molar Mass
particlesA
volume of gas A(liters)
molesB
Use balanced equation
1 Mole = 22.4 L of gas at STP
molesA
mass B(grams)
Use Molar Mass
volume of gas B(liters)
1 Mole = 22.4 L of gas at STP
Use balanced equation
How many molecules of O2 are produced when 40.8 moles of KClO3 decompose?
2KClO3 → 2KCl + 3O2
Step 1: Put what you start with over 1.
40.8 moles KClO3
1
Step 2: Draw a multiplication sign and another line.
x
40.8 moles KClO3
1
Step 3: Write the units you want to cancel out in the denominator of this new fraction.
moles KClO3
x
40.8 moles KClO3
1
moles O2
One mole of KClO3 has a mass of 122.6 grams.
Step 4: Write the units of your next stop in the numerator.
moles KClO3
x
40.8 moles KClO3
1
Step 5: Use the corresponding conversion factor to write in the values.
3 moles O2
2 moles KClO3
x
40.8 moles KClO3
1
The balanced reaction shows the ratio of O2 to KClO3 is 3:2.
2KClO3 → 2KCl + 3O2
Repeat Step 2: Draw a multiplication sign and another line.
x
2 mole KClO3
1 mole KClO3
x
122.6 g KClO3
44.8 g KClO3
1
x
3 mole O2
3 moles O2
2 moles KClO3
x
40.8 moles KClO3
1
= 3.68 x 1025 molecules O2
Step 6: Cancel out units that are in both the numerator and denominator. Then, multiply the top numbers and divide by the bottom numbers to get your answer!
x
3 moles O2
2 moles KClO3
x
40.8 moles KClO3
1
Repeat Step 5: Use the corresponding conversion factor to write in the values.
1 mole O2
x
6.02 x 1023 molecules O2
1 mole of O2 contains 6.02 x 1023 molecules.
3 moles O2
2 moles KClO3
x
40.8 moles KClO3
1
Repeat Step 4: Write the units of your next stop in the numerator.
mole O2
x
molecules O2
3 moles O2
2 moles KClO3
x
40.8 moles KClO3
1
mole O2
Repeat Step 3: Write the units you want to cancel out in the denominator of this new fraction.
x
3 moles O2
2 moles KClO3
x
40.8 moles KClO3
1
1 mole O2
6.02 x 1023 molecules O2
Use balanced equation
Next
1 Mole = 6.02 x 1023 particles
particlesB
1 Mole = 6.02 x 1023 particles
mass A(grams)
Use Molar Mass
particlesA
volume of gas A(liters)
molesB
Use balanced equation
1 Mole = 22.4 L of gas at STP
molesA
mass B(grams)
Use Molar Mass
volume of gas B(liters)
1 Mole = 22.4 L of gas at STP
Use balanced equation
How many grams of KCl are produced when 36.5 moles of KClO3 decompose?
2KClO3 → 2KCl + 3O2
Step 1: Put what you start with over 1.
36.5 moles KClO3
1
Step 2: Draw a multiplication sign and another line.
x
36.5 moles KClO3
1
Step 3: Write the units you want to cancel out in the denominator of this new fraction.
moles KClO3
x
36.5 moles KClO3
1
moles KCl
One mole of KClO3 has a mass of 122.6 grams.
Step 4: Write the units of your next stop in the numerator.
moles KClO3
x
36.5 moles KClO3
1
Step 5: Use the corresponding conversion factor to write in the values.
2 moles KCl
2 moles KClO3
x
36.5 moles KClO3
1
The balanced reaction shows the ratio of KCl to KClO3 is 2:2.
2KClO3 → 2KCl + 3O2
Repeat Step 2: Draw a multiplication sign and another line.
x
2 mole KClO3
1 mole KClO3
x
122.6 g KClO3
44.8 g KClO3
1
x
3 mole O2
2 moles KCl
2 moles KClO3
x
36.5 moles KClO3
1
mole KCl
Repeat Step 3: Write the units you want to cancel out in the denominator of this new fraction.
x
2 moles KCl
2 moles KClO3
x
36.5 moles KClO3
1
Repeat Step 4: Write the units of your next stop in the numerator.
mole KCl
x
g KCl
2 moles KCl
2 moles KClO3
x
36.5 moles KClO3
1
Repeat Step 5: Use the corresponding conversion factor to write in the values.
1 mole KCl
x
74.5 g KCl
1 mole of KCl has a mass of 74.5 grams.
2 moles KCl
2 moles KClO3
x
36.5 moles KClO3
1
= 2720 g KCl
Step 6: Cancel out units that are in both the numerator and denominator. Then, multiply the top numbers and divide by the bottom numbers to get your answer!
1 mole KCl
x
74.5 g KCl
2 moles KCl
2 moles KClO3
x
36.5 moles KClO3
1
For one set of cards (cards=pictures/text/object that you want to appear)
You need the orange element on every page. It will also display errors (in Spanish)This element shows either as "baraja1" or as "set1". Copy and paste to create as many as you have cards. Group the element with your cards IN THE ORDER that you want them to appear. Go to a differrent slide and return to see the updated numbering. The numbering changes with every times you group/ungroup or duplicate elements.Optional Elements:Group the "contador" with any text. The text will be automatically replaced to display how many rounds have been played.Group the "Terminado" element with text/object that you want to appear once all cards have been used.Group the "Reiniciar" element with a button (e.g. an image or text) . When the button is pressed, the game is reset.Group the "Ocultar" with a text/object that you want to disappear once the game starts, e.g. the instructionsChoose one type of these buttons to decide how cards will be drawn:
display a random element with animation.
display a new element, hiding the rest
display a new element, keeping the rest visible
to make the last two buttons random, add the black box to the page. Only group with "true" (for random) or "false" (for not random)
true
For several sets of cards
This page allows for up to 5 decks of cards. See next page if you need more.For each set of cards, use a different set of "baraja/set" elements. Each cards of deck 1 need to be grouped with a "baraja/set 1" button, each card of deck 2 need to be grouped with a "baraja/set 2" button etc. Also use a different set of set of buttons from the white box for each set (Ocultar 1 for the first set, Ocultar 2 for the second set etc.)
Group the elements from the orange box with buttons that will apply to ALL sets of cards(e.g. it draw the next card in all sets at the same time or it will reset all cards at the same time)
Elements for up to 10 sets If you need more sets, use the codes on the next 2 pages. Change the numbers in bolt to create more elements. You can also change the colour names in the codes to give it different colours to be able to tell them apart easier. Once the code is changed, go to "insert", then "other" and paste it in the box to create the element.
<div class="marcadorBaraja11" id="marcadorBaraja11" style="font-size:16pt;">marcadorBaraja11</div> <script> </script>
<div class="marcadorBaraja2" id="marcadorBaraja2" style="font-size:16pt;">marcadorBaraja2</div> <script> </script>
<div class="marcadorBaraja3" id="marcadorBaraja3" style="font-size:16pt;">marcadorBaraja3</div> <script> </script>
<div class="marcadorBaraja4" id="marcadorBaraja4" style="font-size:16pt;">marcadorBaraja4</div> <script> </script>
<div class="marcadorBaraja5" id="marcadorBaraja5" style="font-size:16pt;">marcadorBaraja5</div> <script> </script>
<div class="mostrarCartasAll" style="opacity:0.5; color:black; background-color:yellowgreen; text-align:center; font-weight:normal; font-size:16px; width:100%; height:100%; ">Mostrar Nueva ALL </div> <script> </script>
<div class="reiniciarPistasALL" style="opacity:0.5; color:black; background-color:hotpink; text-align:center; font-weight:normal; font-size:16px; width:100%; height:100%; ">Reiniciar Pistas ALL </div> <script> </script>
<div class="tiraDadoALL" style="opacity:0.5; color:black; background-color:palegreen; text-align:center; font-weight:normal; font-size:16px; width:100%; height:100%; ">Tira Dados ALL </div> <script> </script>
<div class="avanzaCartaALL" style="opacity:0.5; color:black; background-color:yellow; text-align:center; font-weight:normal; font-size:16px; width:100%; height:100%; ">Avanza Carta ALL </div> <script> </script>
<div class="ocultarObjeto1" style="opacity:0.75; color:white; background-color:orange; text-align:center; font-weight:normal; font-size:16px; width:100%; height:100%; ">Ocultar al Inicio 1 </div> <script> </script>
<div class="terminado1" style="opacity:0.75; color:white; background-color:red; text-align:center; font-weight:normal; font-size:16px; width:100%; height:100%; "> Terminado 1 </div> <script> </script>
<div class="tiraDado1" style="opacity:0.75; color:white; background-color:darkgreen; text-align:center; font-weight:normal; font-size:16px; width:100%; height:100%; ">Tira dado 1 </div> <script> </script>
<div class="avanzaCartas11" style="opacity:0.75; color:black; background-color:aquamarine; text-align:center; font-weight:normal; font-size:16px; width:100%; height:100%; ">Avanza Nueva 11</div> <script> </script>
<div class="mostrarCartas1" style="opacity:0.75; color:black; background-color:#00ff00; text-align:center; font-weight:normal; font-size:16px; width:100%; height:100%; ">Mostrar Nueva 1 </div> <script> </script>
<div class="reiniciarBaraja1" style="opacity:0.75;color:white; background-color:magenta; text-align:center; font-weight:normal; font-size:16px; width:100%; height:100%; "> Reiniciar baraja 1 </div> <script> </script>
<div class="baraja1" style="font-size:16pt;">baraja1</div> <script> </script>