Stoichiometry Worksheet Page 6
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6Limiting Reactant (LO 3.4)
In a limiting reactant problem you will be given amounts of two of your reactants. The goal is to determine
which you will run out of first. The only way to do this is to solve the stoichiometry problem using each
reactant and its value. The reactant that produces less of the same product is the one you would run out of
first, and thus your limiting reactant. The other reactant, the one you will have extra of when finished, is
your excess reactant.
Percent Yield
The percent yield equation is used to determine the efficiency of a chemical procedure. When you perform
a percent yield calculation you will be comparing how much product you actually produce in a lab
situation with how much should theoretically be produced if the situation was ideal. The theoretical yield
is what you determine in a stoichiometry problem. The percent yield equation is listed below.
Percent Yield = Actual Yield
x 100
Theoretical Yield
An example showing all three topics and the worked out solutions are included below.
Zn + 2HCl ZnCl
.
+ H
2
2
Zinc reacts with hydrochloric acid to form zinc chloride and hydrogen gas. 98.2 grams of zinc and 98.2 grams of
hydrogen chloride react?
(a) Identify the limiting reactant. Support your answer with calculations. (3 points)
(b) How much of the excess reagent remains? (3 points)
(c) Calculate the volume of hydrogen gas produced
(3 points)
.
(d) If 12.5 liters of hydrogen gas are actually produced, what is the percent yield? (3 points)
(a)
98.2 g Zn x 1 mole Zn x 1 mole H
= 1.50 mol H
2
2
1
65.4 g Zn
1 mole Zn
98.2 g HCl x 1 mole HCl x 1 mole H
= 1.34mol H
2
2
1
36.5 g HCl 2 mole HCl
Limiting Reagent: HCl
(b)
98.2 g HCl x 1 mole HCl x 1 mole Zn x 65.4 g Zn = 88.0 g Zn needed
1
36.5 g HCl
2 mole HCl
1 mole Zn
98.2 – 88.0 = 10.2 g of excess Zn
(c)
98.2 g HCl x 1 mole HCl x 1 mole H
x
22.4 L H
= 30.1 L H
2
2
2
1
36.5 g HCl
2 moles HCl
1 mole NH
2
(d)
Percent Yield = 12.5 x 100 = 41.5%
30.1
Balancing Answers:
10 C + 16 HCl
OC(NH
1. C
H
+ 8 Cl
2. CO
+ 2 NH
)
+ H
O
10
16
2
2
3
2
2
2
8 SiO
Al
3. 4 Si
H
+ 11 O
+ 6 H
O
4. 2 Al(OH)
+ 3 H
SO
(SO
)
+ 6 H
O
2
3
2
2
2
3
2
4
2
4
3
2
2 Fe
+ 6 KOH 3 K
5. 4 Fe + 3 O
O
6. Fe
(SO
)
SO
+ 2 Fe(OH)
2
2
3
2
4
3
2
4
3
14 CO
+ 8 HI H
7. 2 C
H
O
+ 15 O
+ 6 H
O
8. H
SO
S + 4 I
+ 4 H
O
7
6
2
2
2
2
2
4
2
2
2
6
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