CHEM107_Exam2_Study Guide Answer - Texas A M University...

Texas A & M University
Department of Chemistry
CHEM107
Spring 2011
Instructor: Masud Monwar, Ph.D.
Study Guide for Exam 2
28th February, 2011
Chapter 4: Stoichiometry
1. What mass of HCl, in grams, is required to react with 0.750 g of Al(OH)
3
. What mass of
H
2
O, in grams, is produced? What mass of AlCl
3
, in grams, is produced?
The equation for the reaction is:
Al(OH)
3
+ HCl -----------  AlCl
3
+ H
2
O
Solution:
Molar mass of Al(OH)
3
= (1 × 27.0) + (3 × (1.0 + 16.0)) = 78.0 g/mol
Molar mass of HCl = (1 × 1.0) + (1 × 35.45) = 36.45 g/mol
Molar mass of H
2
O = (2 × 1.0) + (1 × 16.0) = 18.0 g/mol
Molar mass of AlCl
3
= (1 × 27.0) + (3 × 35.45) = 133.35 g/mol
0.750 g of Al(OH)
3
× = 9.615 × 10 -3 mol of Al(OH)
3
9.615 × 10 -3 mol of Al(OH)
3
× = 0.0288 mol of HCl required
0.0288 mol of HCl × = 1.04976 g = 1.050 g of HCl required
9.615 × 10 -3 mol of Al(OH)
3
× = 9.615 × 10 -3 mol of AlCl
3
produced
Mass of AlCl
3
produced = 9.615 × 10 -3 mol of AlCl
3
× = 1.282 g of AlCl
3
9.615 × 10 -3 mol of Al(OH)
3
× = 0.0288 mol of H
2
O produced
Mass of H
2
O produced = 0.0288 mol of H
2
O × = 0.5184 g of H
2
O
Continued to page 2
2. Consider the following reaction:
P
4
(s) + 6Cl
2
(g) ------------  4PCl
3
(s)
Calculate the mass of phosphorus (P
4
) required to produce 15.0 g of phosphorus trichloride,
PCl
3
?

Solution:
Molar mass of P
4
= (4 × 30.97) = 123.88 g/mol
Molar mass of PCl
3
= (1 × 30.97) + (3 × 35.45) = 137.32 g/mol
Mol of PCl
3
= 15.0 g of PCl
3
× = 0.1092 mol of PCl
3
Mols of P
4
required = 0.1092 mol of PCl
3
× = 0.0273 mol of P
4
required
Mass of P
4
required = 0.0273 mol of P
4
× = 3.382 g of P
4
3. a. What is the molarity of the solution when 2.0 g of KOH is dissolved in 500 mL of
solution?
Solution:
Molar mass of KOH = (39.0 + 16.0 + 1.0) = 56.0 g/mol
2.0 g KOH × = 0.0357 mol
500 mL × = 0.5 L
Molarity = = = 0.071 mol/L or, M
b. What is the mass of solute, in grams, in 250.0 mL of a 0.0125 M solution KMnO
4
?
Solution:
Molar mass of KMnO
4
= (39.0 + 54.94 + 4×16.0) = 157.4 g/mol
250 mL × = 0.25 L
Molarity =
0.0125 mol/L = = 0.071 mol/L or, M
Number of mols = 0.0125 mol/L × 0.25 L = 3.125 × 10 -3 mol
Mass of KMnO
4
= 3.125 × 10 -3 mol × = 0.49 g
Continued to page 3
c. What volume of 2.06 M KMnO
4
, in liters, contains 322.0 g of solute?
Solution:
Molar mass of KMnO
4
= (39.0 + 54.94 + 4×16.0) = 157.4 g/mol
322.0 g of KMnO
4
× = 2.045 mol
Molarity =
2.06 mol/L =
Volume(L) = = 0.9927 L × = 992.7 mL

d. If 4.00 mL of 0.0250 M CuSO
4
is diluted to 10.0 mL with pure water, what is the molar
concentration of copper(II) sulfate in the diluted solution?
Solution:
M1V1 = M2V2
4.00 × 10 -3 L × 0.0250 mol/L = M2 × 10.0 × 10 -3 L
M2 = = 0.01 mol/L or, M
4. What mass of HCl, in grams, is required to react with 0.750 g of Al(OH)
3
. What mass of
H
2
O, in grams, is produced? What mass of AlCl
3
, in grams, is produced?
The equation for the reaction is:
Al(OH)
3
+ HCl -----------  AlCl
3
+ H
2
O
Solution:
Balance the equation: Al(OH)
3
+ 3HCl -----------  AlCl
3
+ 3H
2
O
Molar mass of Al(OH)
3
= 27.0 + (3×17.0) = 78.0 g/mol
Molar mass of HCl = (1.0 + 35.45) = 36.45 g/mol
Molar mass of H
2
O = (2 × 1.0 + 16.0) = 18.0 g/mol
Molar mass of AlCl
3
= (27.0 + 3 × 35.45) = 133.35 g/mol
0.750 g of Al(OH)
3
× = 9.615 × 10
-3
mol of Al(OH)
3
9.615 × 10 -3 mol of Al(OH)
3
× = 0.0288 mol HCl required
9.615 × 10
-3
mol of Al(OH)
3
× = 9.615 × 10
-3
mol of AlCl
3
produced
9.615 × 10 -3 mol of Al(OH)
3
× = 0.0288 mol of H
2
O produced
Continued to page 4
5. Ammonia gas can be prepared by the reaction of a metal oxide such as calcium oxide
with ammonium chloride.