mol x 164.10 g/mol =
1.64 g Ca(NO3)2
b) 1.0 x 10
2
mol Ca(OH)2 x 2 mol HNO3/1 mol Ca(OH)2 =
2 x 10
2
mol of Ca(OH)2
What is left over: 3.00 x 10
2
mol
–
1.0 x 10
2
mol =
2.0 x 10
2
mol HNO3
2. Consider the following reaction:
P (s)
+
S (s)
→
P
4
S
3
(s)
a) How many grams of tetraphosphorus trisulfide
can be produced from 62.0 g of phosphorus and
38.6 g of sulfur? B) How many grams of which reagent will remain unreacted?
4 P (s)
+
3 S (s)
→
P
4
S
3
(s)
a) n = m/M = 62.0 g/30.97 g/mol =
2.00 mol P
n = m/M = 38.6 g/32.07 g/mol
= 1.20 mol S
2.00 mol P x 3 mol S/4 mol P =
1.50 mol S
:
The limiting reagent is
S
.
1.20 mol S x 1 mol P4S3/3 mol S = 0.400 mol P4S8
m = nM = 0.400 mol x 220.06 g/mol =
88.0 g of P4S3
b) Excess reagent: 1.2 mol S x 4 mol P/3 mol S =
1.6 mol P will be used up
Unreacted reagent: 2.00 mol
–
1.6 mol =
0.400 mol of P will remain
.
M = nM = 0.400 mol x 30.97 g/mol =
12.388 g of P will remain
.
1.11.1 Exercises:
3. Consider the following reaction:
HBr (aq)
+
Fe(OH)
3
(s)
→
FeBr
3
(aq)
+
H
2
O (l)
a)
What weight of iron (III) bromide can be produced by the reaction of 30.0 g of hydrogen bromide
with 10.0 g of iron (III) hydroxide? B) Given that the percentage yield is 66%, how many grams of
iron (III) bromide are obtained?
3 HBr (aq)
+
Fe(OH)
3
(s)
→
FeBr
3
(aq)
+
3 H
2
O (l)
a) HBr:
n = m/M = 30.0 g/80.91 g/mol
= 0.371 mol HBr
Fe(OH)3:
n = m/M = 10.0 g/106.88 g/mol =
0.0936 mol Fe(OH)3
Limiting reagent calculations:
0.371 mol HBr x 1 mol Fe(OH)3/3 mol HBr =
0.124 mol Fe(OH)3
Fe(OH)3 is the limiting reagent.
0.0936 mol Fe(OH)3 x 1 mol FeBr3/1 mol Fe(OH)3 =
0.0936 mol
m = nM = 0.0936 mol x 295.55 g/mol =
27.66 g FeBr3
b) 66% x 27.66 g =
18.26 g
1.12.1 Exercises:
1. A solution can be prepared by mixing 1.00 g of ethanol (C
2
H
5
OH) with 100.0 g of water to give a
final volume of 101 mL. Calculate the molarity of ethanol in the solution.
n = m/M = 1.00 g/46.08 g/mol = 0.0217 mol
M = n/v = 0.0217 mol/0.101 l =
0.215 mol/L
2. A sulphuric acid solution of density 1.802 g/mL contains 88.0% H
2
SO
4
by weight. A) Find the
weight of H
2
SO
4
per liter of solution. B) Find the molarity of the solution.
a) 1 liter = 1 802 g
0.880 x 1 802 g =
1.59 x 10
3
g
b) n = m/M = 1.59 x 10
3
g/98.09 g/mol
= 16.2 mol
M = n/v = 16.2 mol/1 L =
16.2 mol/L
3. A 325 mL sample of a solution contains 25.3 g CaCl
2
(111 g/mol). A) Calculate the molarity of Cl

ions in solution. B) How many grams of Cl

are there in 0.100L of this solution?
a) n = m/M = 25.3 g/110.98 g/mol = 0.2280 mol
M = n/v = 0.2280 mol/0.325 L =
0.702 mol/L
b) n = MV = 0.702 mol/L x 0.100 L = 0.0702 mol
m = nM = 0.0702 mol x 35.45 g/mol =
2.49 g of Cl

.
4. Find the weight percent of calcium carbonate in a 2.00 g sample of rock if it reacts with 30.0 mL of
0.60 M H
2
SO
4
according to the following reaction:
CaCO
3
(s)
+
H
2
SO
4
(aq)
→
CaSO
4
(s)
+
H
2
O
(l)
+
CO
2
(g)
n = MV = 0.60 mol/L x 0.0300 L = 0.0180 mol
m = nM = 0.0180 mol x 100.09 g/mol = 1.80 g
Weight % = 1.80 g/2.00 g x 100% =
90.0%
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 Fall '14
 Chemistry, Sulfuric acid, mol