Stoichiometry Practice Sheet
Init 1/10/2012 by Daniel R. Barnes
WARNING: This presentation may contain graphical images and other content that was taken from the world wide web without permission of the owner(s). Please do not copy or distribute this presentation. Its very existence may be illegal.

Propane combusts in the presence of oxygen (burns) according to the following balanced equation:
C
3
H
8
+ 5O
2
 3CO
2
+ 4H
2
O
1. Draw a cartoon of the reaction below, represeting each atom as a circle with its element’s symbol inside.
H
H H H
C C C
H H H
H
O
O
O
C
O
O
C
O
H H
O
O
C
O
O
O
O
O
O
O
O
O
H H
O
H H
O
H H
O

Propane combusts in the presence of oxygen (burns) according to the following balanced equation:
1 C
3
H
8
+ 5O
2
 3CO
2
+ 4H
2
O molecule burned?
A: five
H
H H H
C C C
H H H
H
Would looking at the cartoon help?
O
O
O
O
O
O
O
O
O
O
O
C
O
O
C
O
O
C
O
H H
O
H H
O
H H
O
H H
O

Propane combusts in the presence of oxygen (burns) according to the following balanced equation:
1 C
3
H
8
+ 5O
2
 3CO
2
+ 4H
2
O
2. How many oxygen molecules are consumed for every propane molecule burned?
A: five Would looking at the cartoon help?
One molecule
H H H of propane
H C C C H
H H H O
4
O
1
O
3
O
O
O
2
O
O
5
O O
Five molecules of oxygen
O
C
O
O
O
H H
H H
O
C
O
H H
O
O
C
H H
O
O

Propane combusts in the presence of oxygen (burns) according to the following balanced equation:
1 C
3
H
8
+ 5O
2
 3CO
2
+ 4H
2
O
3. How many carbon dioxide molecules are produced for every propane molecule burned?
A: three Would looking at the cartoon help?
H
H H H
C C C
H H H
H
O
O
O
O
O
O
O
O
O
O
O
C
O
O
C
O
O
C
O
H H
O
H H
O
H H
O
H H
O

Propane combusts in the presence of oxygen (burns) according to the following balanced equation:
1 C
3
H
8
+ 5O
2
 3CO
2
+ 4H
2
O
3. How many carbon dioxide molecules are produced for every propane molecule burned?
Three molecules of
A: three carbon dioxide
H
H
H
O
O
H
H
O
O
H
C C C
H
H
O
O
One molecule of propane
O
O
O
O
O
1
C
O
O
H H
O
H H
O
2
C
O
H H
O
O
C
3
O
H H
O

Propane combusts in the presence of oxygen (burns) according to the following balanced equation:
C
3
H
8
+ 5O
2
 3CO
2
+ 4H
2
O
4. How many water molecules are created for every oxygen molecule consumed?
4 water molecules
5 oxygen molecules
= 0.8 water molecules/oxygen molecule

Propane combusts in the presence of oxygen (burns) according to the following balanced equation:
C
3
H
8
+ 5O
2
 3CO
2
+ 4H
2
O
5. If two million propane molecules are burned, how many water molecules are produced?
2 million propane molecules x
1
4 water molecules propane molecule
= 8 million water molecules

Propane combusts in the presence of oxygen (burns) according to the following balanced equation:
C
3
H
8
+ 5O
2
 3CO
2
+ 4H
2
O
6. How many oxygen molecules does it take to produce nine billion carbon dioxide molecules?
9 billion carbon dioxide molecules x
1
5 oxygen molecules
3 carbon dioxide molecules
= 15 billion oxygen molecules

Propane combusts in the presence of oxygen (burns) according to the following balanced equation:
C
3
H
8
+ 5O
2
 3CO
2
+ 4H
2
O
7. How many propane molecules does it take to produce 484 dozen water molecules?
484 dozen water molecules x
1
1 propane molecules
4 water molecules
= 121 dozen propane molecules

Propane combusts in the presence of oxygen (burns) according to the following balanced equation:
C
3
H
8
+ 5O
2
 3CO
2
+ 4H
2
O
8. How much carbon dioxide would be produced by burning
10 kazillion oxygen molecules?
10 kazillion oxygen molecules x
1
3 carbon dioxide molecules
5 oxygen molecules
= 6 kazillion carbon dioxide molecules

Propane combusts in the presence of oxygen (burns) according to the following balanced equation:
C
3
H
8
+ 5O
2
 3CO
2
+ 4H
2
O
9. For every mole of carbon dioxide produced, how many moles of water are produced?
1 mol carbon dioxide
1 x
4 mol water
3 mol carbon dioxide
= 1.33 mol water
You could also say, “1.33 mol water/mol carbon dioxide.”

Propane combusts in the presence of oxygen (burns) according to the following balanced equation:
C
3
H
8
+ 5O
2
 3CO
2
+ 4H
2
O
10. What is the oxygen-to-carbon dioxide mole ratio in this reaction?
5 to 3 = 5:3 =
=
3
5 oxygen molecules carbon dioxide molecules
3
5 mol oxygen mol carbon dioxide

Propane combusts in the presence of oxygen (burns) according to the following balanced equation:
C
3
H
8
+ 5O
2
 3CO
2
+ 4H
2
O
11. In general, how do you figure out the mole ratio of any two substances in a chemical reaction?
The coefficients

And now, for the of the worksheet
(Sorry. The rest of the front side is under construction for now.)

Mysterious clue:

12. What mass of magnesium oxide should be produced when
12.15 grams of magnesium metal burns? Magnesium burns according to the following equation: 2Mg + O
2
 2MgO
12.15 g
2Mg + O
2
20
 2MgO g
12.15 g Mg
1
Mg:
Mg: 1 x 24 = 24
1
24 mol Mg g Mg
24 g/mol
2
2 mol MgO mol Mg
MgO:
Mg: 1 x 24 = 24
O: 1 x 16 = 16
40
1 g MgO mol MgO
40 g/mol
12.15 x 2 x 40 g MgO =
24 x 2
960
48 g MgO = 20 g MgO
Just show me the answer!

12. What mass of magnesium oxide should be produced when
12.15 grams of magnesium metal burns? Magnesium burns according to the following equation: 2Mg + O
2
 2MgO
12.15 g
2Mg + O
2
20
 2MgO g
12.15 g Mg
1
Mg:
Mg: 1 x 24 = 24
1
24 mol Mg g Mg
24 g/mol
2
2 mol MgO mol Mg
MgO:
Mg: 1 x 24 = 24
O: 1 x 16 = 16
40
1
40 g/mol
12.15 x 2 x 40 g MgO =
24 x 2
960
48 g MgO = 20 g MgO g MgO mol MgO

13. How many grams of hydrogen gas can 128 grams of oxygen burn? Hydrogen gas burns according to the following balanced equation: 2H
2
+ O
2
 2H
2
O.
16
2H
2 g
128 g
+ O
2
 2H
2
O
128 g O
1
2
O
2
:
O: 2 x 16 = 32
1
32 mol O
2 g O
2
32 g/mol
2
1 mol H mol O
2
2
H
2
:
H: 2 x 1 = 2
2 g/mol
2
1 g H
2 mol H
2
128 x 2 x 2
32 g H
2
=
512
32 g H
2
= 16 g H
2
Just show me the answer!

13. How many grams of hydrogen gas can 128 grams of oxygen burn? Hydrogen gas burns according to the following balanced equation: 2H
2
+ O
2
 2H
2
O.
16
2H
2 g
128 g
+ O
2
 2H
2
O
128 g O
1
2
O
2
:
O: 2 x 16 = 32
1
32 mol O
2 g O
2
32 g/mol
2
1 mol H mol O
2
2
H
2
:
H: 2 x 1 = 2
2 g/mol
2
1 g H
2 mol H
2
128 x 2 x 2
32 g H
2
=
512
32 g H
2
= 16 g H
2

14. What mass of copper (II) oxide should be produced when 620 g copper (II) carbonate decomposes according to the following balanced equation: CuCO
3
 CuO + CO
2
?
620 g
400 g
CuCO
3
 CuO + CO
2
620 g CuCO
1
3
1 mol CuCO
3
124 g CuCO
3
CuCO
3
:
Cu: 1 x 64 = 64
C: 1 x 12 = 12
O: 3 x 16 = 48
124 g/mol
1
1 mol CuO mol CuCO
CuO:
Cu: 1 x 64 = 64
O: 1 x 16 = 16
3
80 g CuO
1 mol CuO
80 g/mol
620 X 80
124 g CuO =
49,600
124 g CuO = 400 g CuO
Just show me the answer!

14. What mass of copper (II) oxide should be produced when 620 g copper (II) carbonate decomposes according to the following balanced equation: CuCO
3
 CuO + CO
2
?
620 g
400 g
CuCO
3
 CuO + CO
2
620 g CuCO
1
3
1 mol CuCO
3
124 g CuCO
3
CuCO
3
:
Cu: 1 x 64 = 64
C: 1 x 12 = 12
O: 3 x 16 = 48
124 g/mol
1
1 mol CuO mol CuCO
CuO:
Cu: 1 x 64 = 64
O: 1 x 16 = 16
3
80 g CuO
1 mol CuO
80 g/mol
620 X 80
124 g CuO =
49,600
124 g CuO = 400 g CuO

15. How many grams of NI
3 must explode in order to produce
7620 grams of iodine vapors? Nitrogen triiodide decomposes according to the following balanced equation: 2NI
3
 N
2
+ 3I
2
7900 g
2NI
3
 N
2
7620 g
+ 3I
2
7620 g I
1
2
I
2
:
I: 2 x 127 = 254
1 mol I
2
254 g I
2
254 g/mol
2
3 mol NI
3 mol I
2
NI
3
:
N: 1 x 14 = 14
I: 3 x 127 = 381
395 g NI
3
1 mol NI
3
395 g/mol
7620 x 2 x 395 g NI
3
254 x 3
=
6,019,800 g NI
3
762
= 7900 g NI
3
Just show me the answer!

15. How many grams of NI
3 must explode in order to produce
7620 grams of iodine vapors? Nitrogen triiodide decomposes according to the following balanced equation: 2NI
3
 N
2
+ 3I
2
7900 g
2NI
3
 N
2
7620 g
+ 3I
2
7620 g I
1
2
I
2
:
I: 2 x 127 = 254
1 mol I
2
254 g I
2
254 g/mol
2
3 mol NI
3 mol I
2
NI
3
:
N: 1 x 14 = 14
I: 3 x 127 = 381
395 g NI
3
1 mol NI
3
395 g/mol
7620 x 2 x 395 g NI
3
254 x 3
=
6,019,800 g NI
3
762
= 7900 g NI
3

from this point forward.

Mole Bay
Gulf of Guacamole
Grammaw’s house
Particletown

Q: If given g O2 consumed in an alcohol fire, how would you determine L CO2 @ STP produced?
Use molar mass of O2 to convert g O2 to mol O2.
Use coefficients from bal eq to convert mol O2 to mol CO2.
Use “22.4 L/mol” to convert mol CO2 to L CO2 @ STP

If given L H2 gas @ STP, how would you determine how many grams of H2O should be produced by burning it?
Use “22.4 L/mol” to convert L H
2 to mol H2.
Use coefficients from bal eq to convert mol H2 to mol H2O.
Use molar mass of H2O to convert mol H2O to g H2O.

If given grams of solid Fe, how would you determine
# of formula units Fe2O3 produced during the oxidation process?
Use molar mass of Fe to convert g Fe to mol Fe.
Use coefficients from bal eq to convert mol Fe to mol Fe2O3.
Use NA to convert mol Fe2O3 to formula units Fe2O3.

This might be a good time to try
Practice Problem #’s 15 – 20 on pages 364 – 366 in section 12.2 of the textbook.
Get a head start in class and do the rest for homework tonight.

(From the Q2 BMK, given 1/28-29/2014, by Action Learning Systems)
38. Iron reacts with oxygen to form rust,
4Fe (s) + 3O
2
(g)  2Fe
2
O
3
(s).
When 56 g of iron (Fe) reacts with 96 g of oxygen (O
2
), how much rust (Fe
2
O
3
) will form?
This is a limiting reactant stoichiometry problem. It’s stoichiometry because you are given grams of one chemical and asked to determine grams of some other chemical. It’s a limiting reactant problem because you are actually given grams for TWO chemicals, iron and oxygen.
The way to solve this is to do two stoichiometry calculations.
First, calculate how much rust can be made from 56 grams of iron.
Then, calculate how much rust can be made from 96 grams of oxygen. The answer is the lesser of the two results, since oxygen and iron must cooperate to make rust, and a chain is only as strong as its weakest link.

(From the Q2 BMK, given 1/28-29/2014, by Action Learning Systems)
38. Iron reacts with oxygen to form rust,
4Fe (s) + 3O
2
(g)  2Fe
2
O
3
(s).
When 56 g of iron (Fe) reacts with 96 g of oxygen (O
2
), how much rust (Fe
2
O
3
) will form?
Limiting reactant
56 g Fe
1
1
56 mol Fe g Fe
2
4 mol Fe
2
O mol Fe
3
160 g Fe
2
O
3
1 mol Fe
2
O
3
= 80 g Fe
2
O
3
Excess reactant
96 g O
2
1
1 mol O
2
32 g O
2
2
3 mol Fe
2 mol O
2
O
3
160 g Fe
2
O
3
1 mol Fe
2
O
3
= 320 g Fe
2
O
3

If you think there’s something wrong or missing, please E-mail me!
barnesd@centinela.k12.ca.us

#1
#2
TM
#5
Click a button. Go to a place.
#9
#12
Q2 BMK #38 bonus question
#6
#10
#13
#3
#7
#11
#4
#8
#14
#15