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Stoichiometry Practice Sheet
Init 1/10/2012 by Daniel R. Barnes
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Propane combusts in the presence of oxygen (burns) according to the following balanced equation:
C3H8 + 5O2 3CO2 + 4H2O
1. Draw a cartoon of the reaction below, represeting each atom as a circle with its element’s symbol inside.
C C CHH
H
H
H
H
H
H
OO O
O
OO O
O
OO
OC
O
OC
OO
CO
OH H
OH H
OH H
OH H
1
Propane combusts in the presence of oxygen (burns) according to the following balanced equation:
C3H8 + 5O2 3CO2 + 4H2O
2. How many oxygen molecules are consumed for every propane molecule burned?
used up
A: five Would looking at the cartoon help?
C C CHH
H
H
H
H
H
H
OO O
O
OO O
O
OO
OC
OO
CO
OC
O
OH H
OH H
OH H
OH H
1
Propane combusts in the presence of oxygen (burns) according to the following balanced equation:
C3H8 + 5O2 3CO2 + 4H2O
2. How many oxygen molecules are consumed for every propane molecule burned?A: five Would looking at the cartoon help?
C C CHH
H
H
H
H
H
H
OO O
O
OO O
O
OO
OC
OO
CO
OC
O
OH H
OH H
OH H
OH H
One molecule of propane
Five molecules of oxygen
1
2
3
4
5
1
Propane combusts in the presence of oxygen (burns) according to the following balanced equation:
C3H8 + 5O2 3CO2 + 4H2O
3. How many carbon dioxide molecules are produced for every propane molecule burned?A: three Would looking at the cartoon help?
C C CHH
H
H
H
H
H
H
OO O
O
OO O
O
OO
OC
OO
CO
OC
O
OH H
OH H
OH H
OH H
1
Propane combusts in the presence of oxygen (burns) according to the following balanced equation:
C3H8 + 5O2 3CO2 + 4H2O
3. How many carbon dioxide molecules are produced for every propane molecule burned?A: three
C C CHH
H
H
H
H
H
H
OO O
O
OO O
O
OO
OC
OO
CO
OC
O
OH H
OH H
OH H
OH H
One molecule of propane
Three molecules of carbon dioxide
1 2 3
Propane combusts in the presence of oxygen (burns) according to the following balanced equation:
C3H8 + 5O2 3CO2 + 4H2O
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:
C3H8 + 5O2 3CO2 + 4H2O
2 million propane molecules
5. If two million propane molecules are burned, how many water molecules are produced?
water molecules
propane moleculex
1
= 8 million water molecules
4
Propane combusts in the presence of oxygen (burns) according to the following balanced equation:
C3H8 + 5O2 3CO2 + 4H2O
9 billion carbon dioxide molecules
6. How many oxygen molecules does it take to produce nine billion carbon dioxide molecules?
oxygen molecules
carbon dioxide moleculesx
1
= 15 billion oxygen molecules
5
3
Propane combusts in the presence of oxygen (burns) according to the following balanced equation:
C3H8 + 5O2 3CO2 + 4H2O
484 dozen water molecules
7. How many propane molecules does it take to produce 484 dozen water molecules?
propane molecules
water moleculesx
1
= 121 dozen propane molecules
1
4
Propane combusts in the presence of oxygen (burns) according to the following balanced equation:
C3H8 + 5O2 3CO2 + 4H2O
10 kazillion oxygen molecules
8. How much carbon dioxide would be produced by burning 10 kazillion oxygen molecules?
carbon dioxide molecules
oxygen moleculesx
1
= 6 kazillion carbon dioxide molecules
3
5
Propane combusts in the presence of oxygen (burns) according to the following balanced equation:
C3H8 + 5O2 3CO2 + 4H2O
1 mol carbon dioxide
9. For every mole of carbon dioxide produced, how many moles of water are produced?
mol water
mol carbon dioxidex
1
= 1.33 mol water
4
3
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:
C3H8 + 5O2 3CO2 + 4H2O
10. What is the oxygen-to-carbon dioxide mole ratio in this reaction?
=5 to 3 5:3 = carbon dioxide moleculesoxygen molecules
35
=mol carbon dioxide
mol oxygen3
5
Propane combusts in the presence of oxygen (burns) according to the following balanced equation:
C3H8 + 5O2 3CO2 + 4H2O
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:
12.15 g g
12.15 g Mg
1 g Mg
mol Mg
Mg:
Mg: 1 x 24 = 24
24 g/mol
24
1
mol Mg
mol MgO2
2 mol MgO
g MgO
MgO:
Mg: 1 x 24 = 24
40 g/mol
O: 1 x 16 = 16
40
1
2Mg + O2 2MgO
12.15 x 2 x 40
24 x 2g MgO =
960
48g MgO = 20 g MgO
20
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 + O2 2MgO
2Mg + O2 2MgO
g MgO
12.15 g g
12.15 g Mg
1 g Mg
mol Mg
Mg:
Mg: 1 x 24 = 24
24 g/mol
24
1
mol Mg
mol MgO2
2 mol MgO
MgO:
Mg: 1 x 24 = 24
40 g/mol
O: 1 x 16 = 16
40
1
12.15 x 2 x 40
24 x 2g MgO =
960
48g MgO = 20 g MgO
20
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 + O2 2MgO
13. How many grams of hydrogen gas can 128 grams of oxygen burn? Hydrogen gas burns according to the following balanced equation: 2H2 + O2 2H2O.
g
2H2 + O2 2H2O128 g
128 g O2
1 g O2
mol O2
O2:
O: 2 x 16 = 32
32 g/mol
32
1
mol O2
mol H22
1 mol H2
g H2
H2:
H: 2 x 1 = 2
2 g/mol
2
1
128 x 2 x 2
32g H2 =
512
32g H2 = 16 g H2
16
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: 2H2 + O2 2H2O.
g
2H2 + O2 2H2O128 g
128 g O2
1 g O2
mol O2
O2:
O: 2 x 16 = 32
32 g/mol
32
1
mol O2
mol H22
1 mol H2
g H2
H2:
H: 2 x 1 = 2
2 g/mol
2
1
128 x 2 x 2
32g H2 =
512
32g H2 = 16 g H2
16
14. What mass of copper (II) oxide should be produced when 620 g copper (II) carbonate decomposes according to the following balanced equation: CuCO3 CuO + CO2 ?
g
CuCO3 CuO + CO2
620 g
620 g CuCO3
1 g CuCO3
mol CuCO3
CuCO3:Cu: 1 x 64 = 64
124 g/mol
124
1
mol CuCO3
mol CuO1
1 mol CuO
g CuO
CuO:
Cu: 1 x 64 = 64
80 g/mol
80
1
620 X 80
124g CuO =
49,600
124g CuO = 400 g CuO
400
Just show me the answer!
C: 1 x 12 = 12O: 3 x 16 = 48
O: 1 x 16 = 16
14. What mass of copper (II) oxide should be produced when 620 g copper (II) carbonate decomposes according to the following balanced equation: CuCO3 CuO + CO2 ?
g
CuCO3 CuO + CO2
620 g
620 g CuCO3
1 g CuCO3
mol CuCO3
CuCO3:Cu: 1 x 64 = 64
124 g/mol
124
1
mol CuCO3
mol CuO1
1 mol CuO
g CuO
CuO:
Cu: 1 x 64 = 64
80 g/mol
80
1
620 X 80
124g CuO =
49,600
124g CuO = 400 g CuO
400
C: 1 x 12 = 12O: 3 x 16 = 48
O: 1 x 16 = 16
15. How many grams of NI3 must explode in order to produce 7620 grams of iodine vapors? Nitrogen triiodide decomposes according to the following balanced equation: 2NI3 N2 + 3I2
g
2NI3 N2 + 3I2
7620 g
7620 g I2
1 g I2
mol I2
I2:I: 2 x 127 = 254
254 g/mol
254
1
mol I2
mol NI32
3 mol NI3
g NI3
NI3:
N: 1 x 14 = 14
395 g/mol
395
1
7620 x 2 x 395
254 x 3g NI3 =
6,019,800
762g NI3 = 7900 g NI3
7900
Just show me the answer!
I: 3 x 127 = 381
15. How many grams of NI3 must explode in order to produce 7620 grams of iodine vapors? Nitrogen triiodide decomposes according to the following balanced equation: 2NI3 N2 + 3I2
g
2NI3 N2 + 3I2
7620 g
7620 g I2
1 g I2
mol I2
I2:I: 2 x 127 = 254
254 g/mol
254
1
mol I2
mol NI32
3 mol NI3
g NI3
NI3:
N: 1 x 14 = 14
395 g/mol
395
1
7620 x 2 x 395
254 x 3g NI3 =
6,019,800
762g NI3 = 7900 g NI3
7900
I: 3 x 127 = 381
Honors Onlyfrom this point forward.
Stoichiometry
ItinerarySketches
Grammaw’shouse
Mole BayGulf of Guacamole
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 H2 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) + 3O2 (g) 2Fe2O3 (s).When 56 g of iron (Fe) reacts with 96 g of oxygen (O2), how much
rust (Fe2O3) 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) + 3O2 (g) 2Fe2O3 (s).When 56 g of iron (Fe) reacts with 96 g of oxygen (O2), how much
rust (Fe2O3) will form?
56 g Fe
1 g Fe
mol Fe
56
1
mol Fe
mol Fe2O32
4
160
1
g Fe2O3
mol Fe2O3
96 g O2
1 g O2
mol O2
32
1
mol O2
mol Fe2O32
3
160
1
g Fe2O3
mol Fe2O3
= 80 g Fe2O3
= 320 g Fe2O3
Limiting reactant
Excess reactant
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Q2 BMK #38 bonus question
