Chem 2 AP Homework #3-6: Problems pg. 109-110 #3.79, 3.82, 3.83, 3.84, 3.88, 3.92, 3.94, 3.108
3.79
Define limiting reagent and excess reagent.
Limiting reagent is the reactant that is depleted; excess reagent is a reagent that is not depleted.
What is the significance of the limiting reagent in predicting the amount of the product
obtained in a reaction? In a reaction, since the reaction stops when one reactant is depleted, the
amount of limiting reagent determines the amount of each product possible.
Can there be a limiting reagent if only one reactant is present? If there is only one reactant,
then technically it must be the limiting reagent since it will be depleted.
3.82
N2+ 3H2 → 2NH3 From diagram, one sees that
you have 3 molecules of N2 and 10 molecules of H2
Using an “ICF” table helps to answer all questions:
N2
3 H2 →
2 NH3
+
Initial
3 mc
10 mc
0 mc
Change
−3 mc
−9 mc
+6 mc
Final
0 mc
1 mc
6 mc
Finding limiting reagent: The number of N2
molecules shown in the diagram is 3. The balanced equation shows 3 moles H2 ! 1 mole
N2. Therefore, we need 9 molecules of H2 to react completely with 3 molecules of N2.
There are 10 molecules of H2 present in the diagram. H2 is in excess.
What is the limiting reagent: N2 is the limiting reagent because it is completely used up and
there is excess H2
Number of moles of product produced: The mole ratio between N2 and NH3 is 1:2. When 3
molecules of N2 react, 6 molecules of NH3 will be produced.
Number of moles of excess reagent left: 9 molecules of H2 will react with 3 molecules of N2,
leaving 1 molecule of H2 in excess.
3.83
Nitric oxide reacts with oxygen gas to form nitrogen dioxide, a dark-brown gas:
2NO(g) + O2(g) → 2NO2(g)
In one experiment 0.886 mole of NO is mixed with 0.503 mole of O2. Calculate which of the
two reactants is the limiting reagent. Calculate also the number of moles of NO2 produced.
2 mol NO 2
0.886 mol NO ×
= 0.886 mol NO 2
2 mol NO
2 mol NO 2
= 1.01 mol NO 2
1 mol O 2
NO is the limiting reagent; it limits the amount of product produced. The amount of product
produced is 0.886 mole NO2.
Here is an ICF table, to see the information in an organized fashion:
0.503 mol O 2 ×
Initial
Change
Final
2 NO
0.866 mol
−0.866 mol
0 mol
+
O2
0.503 mol
−0.433 mol
0.070 mol
→
2 NO2
0 mol
+0.866 mol
0.866 mol
2
3.84
HOMEWORK #3-6 ANSWER KEY
The depletion of the ozone in the stratosphere has been a matter of great concern among
scientists in recent years. O3 + NO à O2 + NO2
If 0.740 g of O3 reacts with 0.679 g of
NO…
How many grams of NO2 produced?
? mol NO 2 = 0.740 g O3 ×
1 mol O3 1 mol NO 2
×
= 0.0154 mol NO 2
48.00 g O3 1 mol O3
? mol NO 2 = 0.670 g NO ×
1 mol NO 1 mol NO 2
×
= 0.0223 mol NO 2
30.01 g NO 1 mol NO
The initial amount of O3 limits the amount of product that can be formed; therefore, it is the
limiting reagent.
46.01 g NO 2
? g NO2 = 0.0154 mol NO 2 ×
= 0.709 g NO2
1 mol NO 2
How many moles of excess reagent remain?
mol NO reacted = 0.0154 mol NO 2 ×
mol NO initial = 0.670 g NO ×
1 mol NO
= 0.0154 mol NO reacted
1 mol NO 2
1 mol NO
= 0.0223 mol NO
30.01 g NO
mol NO remaining = mol NO initial − mol NO reacted.
mol NO remaining = 0.0223 mol NO − 0.0154 mol NO = 0.0069 mol NO
3.88
Why is the actual yield of a reaction almost always smaller than the theoretical yield?
The actual yield of a reaction is usually less than the theoretical yield because not all reactants
form products (reversible reactions or side products), and not all product may be recovered.
3.92
Ethylene, can be prepared by heating hexane at 800°C: C6H6 à C2H4 + other products
If the yield of ethylene production is 42.5 percent, what mass of hexane must be reacted to
produce 481 g of ethylene?
% yield =
actual yield
× 100%
theoretical yield
42.5% yield =
481 g C2 H 4
× 100%
theoretical yield
theoretical yield C2H4 = 1.13 × 103 g C2H4
The mass of hexane that must be reacted is:
1.13 × 103 g C2 H 4 ×
1 mol C2 H 4 1 mol C6 H14 86.15 g C6 H14
×
×
= 3.47 × 103 g C 6 H14
28.05 g C2 H 4 1 mol C2 H 4
1 mol C6 H14
HOMEWORK #3-6 ANSWER KEY
3.94
Disulfide dichloride is used in the vulcanization of rubber. It is prepared by heating sulfur
in an atmosphere of chlorine:
S8(l) + 4Cl2(g) → 4S2Cl2(l)
What is the theoretical yield of S2Cl2 when 4.06 g of S8 are heated with 6.24g of Cl2
4.06 g S8 ×
1 mol S8
4 mol S2Cl2
×
= 0.0633 mol S2Cl2
256.6 g S8
1 mol S8
6.24 g Cl2 ×
1 mol Cl2
4 mol S2Cl2
×
= 0.0880 mol S2Cl2
70.90 g Cl2
4 mol Cl2
? g S2Cl2 = 0.0633 mol S2Cl2 ×
135.04 g S2Cl2
= 8.55 g S 2Cl 2
1 mol S2Cl2
If the actual yield of S2Cl2 is 6.55 g, what is the percent yield?
actual yield
6.55 g
% yield =
× 100% =
× 100% = 76.6%
theoretical yield
8.55 g
3.108 Fe2O3 + 3 CO à 2 Fe + 3 CO2
One obtains 1.64 × 103 kg Fe from a 2.62 x 103 kg sample of Fe2O3. What is % purity of
sample?
ONE METHOD: Find mass of pure Fe2O3 needed to get amount of Fe obtained.
1.64 × 103 kg Fe ×
1 kg-mol Fe 1 kg-mol Fe 2O3 159.7 kg Fe 2O3
×
×
= 2.34 × 103 kg Fe 2O3 (pure)
55.85 kg Fe
2 kg-mol Fe
1 kg-mol Fe 2O3
percent purity of sample =
percent purity =
grams of pure Fe 2O3
× 100%
grams of Fe 2O3 sample
2.34 × 103 kg pure Fe 2O3
× 100% = 89.6% = purity of Fe 2O3
2.62 × 103 kg sample of Fe 2O3
3
4
HOMEWORK #3-6 ANSWER KEY
ANOTHER METHOD: Calculate kg of Fe that theoretically could be produced if the Fe2O3
sample was pure.
We find the theoretical yield of Fe :
3
2.62 × 10 kg Fe 2O3 ×
1 kg-mol Fe 2O3
2 kg-mol Fe
55.85 kg Fe
3
×
×
= 1.83× 10 kg Fe
159.7 kg Fe 2O3 1 kg-mol Fe 2O3 1 kgmol Fe
percent yield =
actual yield
× 100%
theoretical yield
3
1.64 × 10 kg Fe
percent yield =
× 100% = 89.6% = purity of Fe 2O3
1.83 × 103 kg Fe