Question 10.4.2: Calculate the amount of gas produced when collected by water......
Calculate the amount of gas produced when collected by water displacement.
Aluminum reacts with strong acids such as HCl to form hydrogen gas.
2 Al(s) + 6 HCl(aq) → 3 H_{2}(g) + 2 AlCl_{3}(aq)
In one experiment, a sample of Al reacts with excess HCl and the gas produced is collected by water displacement. The gas sample has a temperature of 22.0 ºC, a volume of 27.58 mL, and a pressure of 738 mm Hg. Calculate the amount of hydrogen gas produced in the reaction.
You are asked to calculate the amount of gas produced in a chemical reaction when it is collected over water at a given pressure and temperature.
You are given the volume, pressure, and temperature of the gas produced in the reaction.
The gas collected is a mixture containing water vapor, which has a vapor pressure of 19.83 mm Hg at 22.0 ºC (Table 10.4.1). Subtract the water vapor pressure from the total pressure of the mixture to calculate the partial pressure of the hydrogen gas in the mixture.
P_{H_{2}} = P_{total} – P_{H_{2}O} = 739 mm Hg – 19.83 mm Hg = 718 mm Hg
Use the ideal gas law to calculate the amount of gas produced in the reaction.
P = 718 mm Hg × \frac{1\text{ atm}}{760\text{ mm Hg}} = 0.945 atm
V = 27.58 mL × \frac{1\text{ L}}{1000\text{ mL}} = 0.02758 L
T = (22.0 + 273.15) K = 295.2 K
n_{H_{2}} = \frac{PV}{RT} = \frac{\left(0.945\text{ atm}\right)\left(0.02758\text{ L}\right) }{\left(0.082057\text{ L} \cdot \text{atm/K} \cdot \text{mol}\right) \left(295.2\text{ K}\right)} = 1.08 × 10^{-3} mol
Table 10.4.1 Vapor Pressure of Water
| Temperature (ºC) | Vapor pressure of \pmb{H_{2}O} (mm Hg) |
| 19 | 16.48 |
| 20 | 17.54 |
| 21 | 18.65 |
| 22 | 19.83 |
| 23 | 21.07 |
| 24 | 22.38 |
| 25 | 23.76 |
| 26 | 25.21 |
| 27 | 26.74 |
| 28 | 28.35 |