Question 6.13: Collecting a Gas over a Liquid (Water) In the following reac...
Collecting a Gas over a Liquid (Water)
In the following reaction, 81.2 mL of O_2 (g) is collected over water at 23 °C and barometric pressure 751 mmHg. What mass of Ag_{2} O(s) decomposed? (The vapor pressure of water at 23 °C is 21.1 mmHg.)
2 Ag_2 O(s) \longrightarrow 4 Ag(s) + O_2 (g)
Analyze
The key concept is that the gas collected is wet, that is, a mixture of O_2 (g) and water vapor. Use P_{tot} = P_{O_2} + P_{H_2 O} to calculate PO_2, and then use the ideal gas equation to calculate the number of moles of O_2. The following conversions are used to complete the calculation: mol O_2 \longrightarrow mol Ag_2 O \longrightarrow g Ag_2 O.
P_{O_2} = P_{bar.} – P_{H_2 O} = 751 mmHg – 21.1 mmHg = 730 mmHg
P_{ O _2} = 730 mmHg \times \frac{1 atm }{760 mmHg } = 0.961 atm
V = 81.2 mL = 0.0812 L
n= ?
R = 0.08206 atm L mol^{-1} K^{-1}
T = 23 °C + 273 = 296 K
n = \frac{PV}{RT} = \frac{0.961 atm \times 0.0812 L}{0.08206 atm L mol^{-1} K^{-1}} = 0.00321 mol
From the chemical equation we obtain a factor to convert from moles of O_2 to moles of Ag_2 O. The molar mass of Ag_2 O provides the final factor.
? g Ag _2 O = 0.00321 mol O _2 \times \frac{2 mol Ag _2 O }{1 mol O _2} \times \frac{231.7 g Ag _2O }{1 mol Ag _2 O } = 1.49 g Ag _2 O
Assess
The determination of the number of moles of O_2 in the sample is the key calculation. We can quickly estimate the number of moles of O_2 in the sample by using the fact that for typical conditions (T ≈ 298 K, P ≈ 760 mmHg), the molar volume of an ideal gas is about 24 L. The number of moles of gas (mostly O_2) in the sample is approximately 0.08 L/24 L ≈ 0.003 mol. This estimate is quite close to the value calculated above.