Page 182 - Theory and Problems of BEGINNING CHEMISTRY

CHAP. 11] MOLARITY 171

STOICHIOMETRY IN SOLUTION
11.25. What mass of H 2 C 2 O 4 can react with 35.0 mL of 1.50 M KMnO 4 according to the following equation?

5H 2 C 2 O 4 + 2 KMnO 4 + 3H 2 SO 4 −→ 2 MnSO 4 + K 2 SO 4 + 10 CO 2 + 8H 2 O
Ans. (0.0350 L)(1.50 mol/L) = 0.0525 mol KMnO 4

5 mol H 2 C 2 O 4 90.0gH C 2 O 4
2
0.0525 mol KMnO 4 = 11.8gH C 2 O 4
2
2 mol KMnO 4 1 mol H 2 C 2 O 4
11.26. Calculate the number of grams of BaSO 4 that can be prepared by treating 35.0 mL of 0.479 M BaCl 2
with excess Na 2 SO 4 .
Ans. BaCl 2 + Na 2 SO 4 −→ BaSO 4 + 2 NaCl
(35.0mL)(0.479 mmol BaCl 2 /mL) = 16.8 mmol BaCl 2

1 mmol BaSO 4 233 mg BaSO 4
16.8 mmol BaCl 2 = 3910 mg = 3.91 g BaSO 4
1 mmol BaCl 2 1 mmol BaSO 4
11.27. When 20.0 mL of 1.71 M AgNO 3 is added to 35.0 mL of 0.444 M CuCl 2 , how many grams of AgCl will
be produced?
+
1.71 mol AgNO 1 mol Ag
Ans. 0.0200 L 3 = 0.0342 mol Ag present
+
1L 1 mol AgNO
3
−
0.444 mol CuCl 2 2 mol Cl
−
0.0350 L = 0.0311 mol Cl present
1L 1 mol CuCl 2
+
−
Ag + Cl −→ AgCl
The chloride is limiting. It will produce 0.0311 mol AgCl.
(0.0311 mol AgCl)(143 g AgCl/1 mol AgCl) = 4.45 g AgCl
11.28. What concentration of NaCl will be produced when 1.00 L of 1.11 M HCl and 250 mL of 4.05 M NaOH
are mixed? Assume that the volume of the final solution is the sum of the two initial volumes.
Ans. (1.00 L)(1.11 mol/L) = 1.11 mol HCl
(0.250 L)(4.05 mol/L) = 1.01 mol NaOH
NaOH + HCl −→ NaCl + H 2 O
So 1.01 mol NaCl will be produced, in 1.25 L:
1.01 mol
= 0.808 M
1.25 L

Supplementary Problems

11.29. Describe in detail how you would prepare 250.0 mL of 4.000 M NaCl solution.
Ans. First, figure out how much NaCl you need:
(0.2500 L)(4.000 mol/L) = 1.000 mol
Since the laboratory balance does not weigh out in moles, convert this quantity to grams:

(1.000 mol NaCl)(58.45 g/mol) = 58.45 g NaCl