Page 204 - Theory and Problems of BEGINNING CHEMISTRY

CHAP. 12] GASES 193

If gas mixtures were involved, their volumes would necessarily be the same and their pressures would be in
the ratios of their numbers of moles.

12.53. What is the ratio of volume of CO 2 produced to O 2 used up, both at the same temperature and pressure, for the
reaction in the prior problem?
Ans. The ratio is2:1,asinthe balanced equation.

12.54. (a) Calculate the initial volume of 3.00 L of gas whose pressure has been increased from 1.00 atm to 2.00 atm.
(b) Calculate the final volume of 3.00 L of gas whose pressure has been increased from 1.00 atm to 2.00 atm.
Ans. Read the problem carefully. What must be assumed in part (a)? in part (b)? Since no mention was made of
temperature, we must assume that the temperature is constant in both parts, or we cannot do the problem.
(a) The initial volume was asked for, so we must assume that 3.00 L is the final volume.
(3.00 L)(2.00 atm)
V 2 P 2
V 1 = = = 6.00 L
P 1 1.00 atm
(b) The final volume was asked for, so 3.00 L must be the initial volume.
V 1 P 1 (3.00 L)(1.00 atm)
V 2 = = = 1.50 L
P 2 2.00 atm
(In each case, doubling the pressure reduced the volume to one-half its initial volume.)

12.55. Calculate the final volume of 1.20 L of gas whose pressure is halved at constant temperature.
Ans. The final volume is 2.40 L. According to Boyle’s law, halving the pressure causes V to be doubled.

12.56. In a certain experiment, when 2.500 g of KClO 3 was heated, some O 2 was driven off. After the experiment,
2.402 g of solid was left. (Not all the KClO 3 decomposed.) (a) Write a balanced chemical equation for the reaction.
(b) What compounds make up the solid? (c) What causes the loss of mass of the sample? (d) Calculate the volume
of oxygen produced at STP.
Ans. (a) 2 KClO 3 −→ 2 KCl + 3O 2
(b) The solid is KCl and unreacted KClO 3 .
(c) The loss of mass is caused solely by the escape of O 2 .

1 mol O 2
(d) 0.098 g O 2 = 0.0031 mol O 2
32.0gO 2
nRT (0.0031 mol)[0.0821 L·atm/(mol·K)](273 K)
V = = = 0.069 L = 69 mL
P 1.00 atm
12.57. Calculate the ratios of volume to Celsius temperature for the data in Table 12-3. Are the ratios the same for all the
temperatures?
Ans. No

12.58. Replot the data of Table 12-1, using P and 1/V . Is the result a straight line? Explain.
Ans. The result is still a straight line. PV = k so P = k(1/V ) and also V = k(1/P).

12.59. What is the final temperature or pressure in each of the following parts: (a) The pressure of a sample of gas at STP
is raised 2 atm. (b) The pressure of a sample of gas at STP is raised to 2 atm. (c) The temperature of a sample of gas
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at STP is raised 20 C. (d) The temperature of a sample of gas at STP is raised to 20 C?
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Ans. (a) 3 atm (b) 2 atm (c)20 C(d)20 C