Honors Chemistry
Name __________________________
Stoichiometry, Limiting Reactant, and
Percent Yield Calculations
Date _____/_____/_____ Period _____
1. How much does 5.3 x 1019 molecules of the glass etching agent hydrofluoric acid weigh?
5.3 x 1019 molecules x
1 mole
6.022 x 1023 molecules
x
20. g HF
1 mole
= 0.0018 g HF
2. Nitrogen monoxide, NO, is an important signaling molecule in cells. It is also a major
component of smog, produced during the high temperature combustion reactions in engines,
according to the unbalanced equation N2 + O2  NO. If an engine completely combusts all
the nitrogen in a 100. g sample of air that is 78.0% nitrogen by mass, with an excess of
oxygen provided by an oxidizing agent added to the gasoline, how many grams of nitrogen
monoxide are produced?
N2
O2  2NO
+
100.g air
78.0 g N2
78.0 g N2
x
excess
1 mole N2 x
28.0 g N2
2 mole NO
1 mole N2
Xg
x
30.0 g NO = 167 g NO
1 mole NO
3. Ideal combustion, resulting in only carbon dioxide and water, rarely happens. In general, at
the very least, some carbon monoxide is also produced. The real combustion of methane is
more closely represented, then, by the unbalanced equation CH4 + O2  H2O + CO2 + CO.
If 0.100 mol CH4 is allowed to react with 3.20 g O2, what is the limiting reagent, if any?
4CH4 + 7O2  8H2O + 2CO2
0.100 mol
0.100 mol CH4
x
3.2 g O2(g)
1 mol O2
32.00 g O2
x
+ 2CO
3.20 g
2 mol CO
4 mol CH4
x
Xg
x
28.01 g CO
1 mol CO
2 mol CO
7 mol O2
x
=
1.40 g CO
28.01 g CO
1 mol CO
= 0.800 g CO
Limiting Reactant = Oxygen gas (yields less product)
4. Using the data from problem three, how many grams of carbon monoxide will be produced?
0.800 g CO (see above)
5. Elemental zinc, Zn, reacts violently with elemental sulfur, S8. If 12.0 g of zinc react with
12.1 x 1024 atoms of sulfur, what is the limiting reagent, if any? (Zinc – see below)
6. Using the data from problem five, how many moles of zinc sulfide, ZnS, are produced?
8Zn +
24
12.0 g
12.0 g Zn
x
1 mol Zn x
65.39 g Zn
12.1 x 1024 atoms S x

S8
12.1 x 10 atoms
8 mol ZnS
8 mol ZnS
1 molecule S8
8 atoms S
x
=
8Zn
X mol
0.184 mol Zn
1 mol S8
x 8 mol ZnS = 20.1 mol
6.022 x 1023 molecules S8
1 mol S8
7. Using the data from problem five, how many grams of excess reactant remains after the
reaction has gone to completion?
Amount of S8 I am using:
12.1 x 1024 atoms S x
1 molecule S8
8 atoms S
x
1 mol S8
x 256.56 g S8 =
23
6.022 x 10 molec. S8
1 mol S8
644 g S8
Amount of S8 that actually reacts:
12.0 g Zn
x
1 mol Zn x 1 mol S8
65.39 g Zn
8 mol ZnS
x 256.56 g S8
1 mol S8
=
5.89 S8 react
Amount of S8 left over (in excess):
644g – 5.89 g = 638.1 g
8. What are two possible reasons for a reaction producing less product(s) than theoretically
calculated?
The reaction did not go to completion (for example – if burning something, maybe not all of the
react was burned)
You may have produced secondary products not considered due to reaction conditions
If the reaction involves water, you may have a “wet product” causing it to reflect more product that
expected?
9. Iron (III) oxide, Fe2O3, reacts with elemental aluminum in a brilliant explosion, forming
elemental iron and aluminum oxide. If 15.0 g of Fe2O3 react with an excess of aluminum,
and careful analysis in a laboratory showed that 7.65 g of Al2O3 is produced, what is percent
yield of the reaction?
Fe2O3 + 2Al  2Fe + Al2O3
15.0 g
15.0 g Fe2O3 x
excess
1 mole Fe2O3 x
160. g Fe2O3
% yield = 7.65 g (actual yield)
9.56 g (expected yield)
7.65 g
1 mole Al2O3 x
1 mole Fe2O3
102 g Al2O3 = 9.56 g
1 mol Al2O3
x 100 = 80.0 %
10. Aqueous solutions of sodium hypochlorite (NaOCl), best known as household bleach are
prepared by the reaction of NaOH with Cl2. How many grams of NaOCl would be produced
if 15.0g of NaOH reacts with 20.0 g of Cl2? What is the limiting reactant? How many
grams of NaOCl will be produced? How many grams of the excess reactant remains after
the reaction has gone to completion? (LR = NaOH)(14.0 g NaOCl produced)(6.7 g excess
Cl2(g) remains)
NaOH(aq)
+ Cl2(aq)

NaOCl(aq)
+
NaCl(aq)
+
H2O()