1.
How many grams of chlorine gas are needed in the production of 200.0 grams of bromine gas from sodium bromide?
Cl
2
+
2 NaBr
®
2 NaCl
+
Br
2
200.0
= 1.2515 mol Br
2
= 1.2515 mol Cl
2
159.808
1.2515 (70.906) = 88.74 g Cl
2
2.
Calculate the percent yield of a reaction that consumes 900.0 grams of potassium chlorate in the preparation of 90.0 grams of oxygen?
2 KClO
3
®
2 KClO
2
+
O
2
900.0
122.548
= 7.344 mol KClO
3
1 O
2
= 3.672 mol O
2
2 KClO
3
3.672 (31.998) = 117.49 g of O
2
(TY)
90.0
(100)
= 76.6% yield
117.49
3.
In the reaction of 84.0 g of copper (II) oxide with hydrogen gas in a reaction with a 92.4% yield, Determine the mass of copper produced?
CuO
+
H
2
®
Cu
+
H
2
O
84.0
= 1.056 mol CuO = 1.056 mol Cu (TY)
79.545
1.056 (63.546) = 67.1 g Cu (TY)
AY
(100) = 92.4 %
67.4
4.
How many grams of sodium hydroxide will react with 1.50 x 10 -3 g of phosphoric acid?
3 NaOH
+
H
3
PO
4
®
3 H
2
AY = 62.0 g of Cu
O
+
Na
3
PO
4
0.00150
97.9937
= 1.53 x 10 -5 mol H
3
PO
4
3 NaOH
PO
4
= 4.59 x 10 -5 mol NaOH
1 H
3
4.59 x 10 -5 (39.9969) = 1.84 x 10 -3 g of NaOH
5.
If 320.0 grams of sodium carbonate react with calcium hydroxide in a 74.67% efficient reaction, how many grams of sodium hydroxide are formed?
Na
2
CO
3
+
Ca ( OH )
2
®
2 NaOH
+
CaCO
3
320.0
105.988
= 3.019 mol Na
2
CO
3
2 NaOH
CO
3
= 6.038 mol NaOH
1 Na
2
6.038 (39.9969) = 241.5 g NaOH (TY)
AY
241.5
(100)
= 74.67% AY = 180.3 g of NaOH

6.
How many grams of sodium iodide are produced by the decomposition of 60.0 g of sodium iodate if the reaction is 89.4% effective?
_ 2 _ NaIO
3
®
_ 2 _ NaI
+
_ 3_ O
2
60.0
= 0.3032 mol NaIO
3
= 0.3032 mol NaI
197.887
0.3032 (149.89) = 45.4466 g NaI (TY)
AY
(100)
=
89.4%
AY = 40.6 g NaI
45.4466
7.
If 10.0 g of aluminum sulfide are produced by the 94.2% effective reaction of aluminum and sulfur, how many grams of sulfur were needed?
_ 2 _ Al
+
_ 3_ S
®
___ Al
2
S
3
10.0
(100)
=
94.2%
TY
10.6
TY = 10.6 g Al
= 0.07059 mol Al
2
S
3
2
S
3
150.159
0.07059
3 S
S
3
0.211775 mol S
Al
2
0.211775 (32.065) = 6.79 g of S
8.
If 2.50 g of cupric sulfide is produced by decomposing cupric sulfate upon heating, how many grams of cupric sulfate was necessary if the reaction is
90.00% efficient?
___ CuSO
4
®
___ CuS
+
_ 2 _ O
2
2.50
(100)
=
90.00%
TY
TY = 2.7777 g of CuS
2.7777
= 0.0290529 mol CuS = 0.0290529 mol CuSO
4
95.611
0.0290529 (159.607) = 4.64 g of CuSO
4
9.
125.0 g of FeS react to form 167.3 g of FeCl
2
: a.
Determine the % yield of the reaction. b.
Calculate, based on that percent yield, the moles of H
2
S that will be produced.
FeS
(s)
+ 2HCl
(aq)
®
FeCl
2(aq)
+ H
2
S
(g)
125.0
= 1.4219 mol FeS = 1.4219 mol FeCl
2
87.910
1.4219 (126.751) = 180.227 g FeCl
2
(TY)

167.3
180.227
(100) = 92.83% yield
10.
If 1.487 moles of lithium oxide react with water through a 79.4% effective reaction, determine the mass of lithium hydroxide produced.
Li
2
O
(s)
+ H
2
O
(l)
®
2LiOH
(aq)
1.487
2 LiOH
1 Li
2
O
= 2.974 mol LiOH
2.974 (23.9479) = 71.22 g LiOH (TY)
AY
71.22
(100)
= 79.4 % AY = 56.5 g LiOH
11.
In order to produce 100.0 g of NaCl based on the following 84.3 % efficient reaction, determine the moles of each reactant required.
Na
2
SO
4(aq)
+ CaCl
2(aq)
®
CaSO
4(s)
+ 2NaCl
(aq)
100.0
58.443
= 1.711 mol NaCl (AY)
1.711
(100)
= 84.3%
TY
TY = 2.0297 mol NaCl
2.0297
1 CaCl
2
2 NaCl
= 1.01 mol CaCl
2
2.0297
1 Na
2
SO
4
2 NaCl
= 1.01 mol Na
2
SO
4
12.
0.941 moles of calcium hydroxide react to form 30.55 g of water. Determine the percent yield of the reaction.
Ca(OH)
2(aq)
+ 2HCl
(aq)
®
CaCl
2(aq)
+ 2H
2
O
(l)
0.941
2 H
2
O
1 Ca ( OH )
2
= 1.882 mol H
2
O
1.882 (18.0148) = 33.9038 g H
2
O (TY)
30.55
33.9038
(100) = 90.11 % yield

13.
What mass of ethane must react in order to produce 4.73 moles of carbon dioxide if the following reaction has a 91.8 % yield?
2C
2
H
6
+ 7O
2
4CO
2
+ 6H
2
O
4.73
(100) = 91.8 %
TY
5.1525
2 C
2
H
6
4 CO
2
TY = 5.1525 mol CO
2
= 2.57625 mol C
2
H
6
2.57625 (30.0694) = 77.5 g of C
2
H
6
14.
4.84 moles of oxygen react in the following 87.9 % yield reaction. Calculate the mass of each product produced.
2Cu
2
S + 3O
2
2Cu
2
O + 2SO
2
4.84
2( Either )
= 3.22666 mol Cu
2
O and SO
2
(TY)
3 O
2
AY
3.22666
(100) = 87.9 % AY = 2.83624 mol of Cu
2
O and SO
2
2.83624 (143.091) = 406 g of Cu
2
O
2.83624 (64.063) = 182 g of SO
2