Stoichiometry Chemistry Worksheets With Answers Page 3
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11oxygen will react completely with 96.1 grams of propane. Notice that for this question, you’ll need to
start by writing the chemical formulas. Now follow these steps:
1. Write the chemical equation.
2. Calculate the molar masses and put them in parentheses above the formulas; soon you’ll
figure out you don’t have to do this for every reactant and product, just those you’re
specifically asked about.
3. Balance the equation.
4. Next put any amounts that you were given into the table. In this example, you were told that
the reaction started with 96.1 g of propane.
5. Find the number of moles of any compounds for which you were given masses. Here you’d
start with propane: you divide 96.1 grams by the molar mass of propane (44.11 g/mol) to get
the number of moles of propane (2.18 mol).
6. Use the mole:mole ratio expressed in the coefficients of each of the compounds to find moles
of all of the necessary compounds involved. The only one you really need to know is oxygen,
but let’s run through all of them for practice. If the coefficient for propane, which is 1, is equal
to 2.18 moles of propane, then the number of moles of oxygen must be 5
2.18 = 10.9, the
moles of CO
is 3
2.18 = 6.54, and the moles of H
O = 4
2.18 = 8.72.
2
2
(44.11)
(32.00)
(44.01)
(18.02)
Molar mass
C
H
+
3CO
+
4H
O
Balanced equation
3
8
2
2
5O
2
No. of moles
2.18
10.9
6.54
8.72
Amount
96.1 g
1. Reread the problem to determine which amount was asked for. The question asks for the
mass of oxygen, so convert moles of oxygen to grams and you have the answer:
10.9 mol
44.01 g/mol = 349 g oxygen
Molar mass
(44.11)
(32.00)
(44.01)
(18.02)
Balanced equation
C
H
+
3CO
+
4H
O
3
8
2
2
5O
2
Mole:mole
1
5
3
4
No. of moles
2.18
10.9
6.54
8.72
Amount
96.1 g
349 g
But what if this question had asked you to determine the liters of CO
consumed in this reaction at
2
STP (273K, 1 atm)? You would take the number of moles of CO
that we calculated from the table and
2
use the standard molar volume for a gas, or 22.4 L/mol. So, 6.54 mol
22.4 L/mol = 146 L.
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