Page 343 - Theory and Problems of BEGINNING CHEMISTRY

332 PRACTICE QUIZZES

Chapter 11

0.100 mmol BaCl 2
1. 225 mL = 22.5 mmol BaCl 2
1mL
0.250 mmol NaCl

208 mL = 52.0 mmol NaCl
1mL
These substances do not react, so we merely add the numbers of millimoles of chloride ion, and divide the number of
millimoles of each ion by the total volume.
−
22.5 mmol BaCl 2 consists of 22.5 mmol Ba 2+ and 45.0 mol Cl .
52.0 mmol NaCl consists of 52.0 mmol Na and 52.0 mmol Cl .
−
+
There is a total of 97.0 mmol of Cl present.
−
The concentration of barium ion is 22.5 mmol/500 mL = 0.0450 M.
The concentration of sodium ion is 52.0 mmol/500 mL = 0.104 M.
The concentration of chloride ion is 97.0 mmol/500 mL = 0.194 M.

0.100 mmol Ba(OH) 2
2. 225 mL = 22.5 mmol Ba(OH) 2
1mL
0.250 mmol HCl

208 mL = 52.0 mmol HCl
1mL
These substances react, so we have a limiting-quantities problem. We use the net ionic equation, and we recognize that
the number of millimoles of barium ion and the number of millimoles of chloride ion will not change.
22.5 mmol Ba(OH) 2 consists of 22.5 mmol Ba 2+ and 45.0 mmol OH .
−
−
+
52.0 mmol HCl consists of 52.0 mmol H and 52.0 mmol Cl .
The concentration of barium ion is 22.5 mmol/500 mL = 0.0450 M.
The concentration of chloride ion is 52.0 mmol/500 mL = 0.104 M.
The hydroxide ion is limiting.

All in millimoles: H + + OH − −→ H 2 O
Initial: 52.0 45.0
Change: 45.0 45.0
Final: 7.0 0.0

The concentration of hydrogen ion is 7.0 mmol/500 mL = 0.014 M.

Chapter 12
nRT (0.0250 mol)(0.0821 L·atm/mol·K)(273 K)
1. V = = = 0.560 L
P (1.00 atm)
2. Note that it is the volume of oxygen gas, not solid KClO 3 that we are to find. The oxygen pressure is the barometric
pressure minus the water vapor pressure:

1 atm
P oxygen = 836 torr − 24 torr = 812 torr = 1.07 atm
760 torr

1 mol KClO 3 3 mol O 2
2.40 g KClO 3 = 0.0295 mol O 2
122 g KClO 3 2 mol KClO 3
nRT (0.0295 mol)(0.0821 L·atm/mol·K)(298 K)
V = = = 0.675 L
P (1.07 atm)