Questions: 13. A sealed tank contains the following mixture of gases: 2.0 grams of helium, 48.0 grams of oxygen, 14.0 grams of nitrogen, 64.0 grams of disulfur trioxide, and 9.0 grams of hydrogen gas. If the total pressure is 800.0 kPa , then find the partial pressure of each gas in atmospheres.

Solution Steps

To find the partial pressure of each gas, we first need to calculate the number of moles of each gas using their respective molar masses.

• Helium (He): Molar mass = 4.0026 g/mol \[ \text{Moles of He} = \frac{2.0 \, \text{g}}{4.0026 \, \text{g/mol}} = 0.4997 \, \text{mol} \]

• Oxygen (O\(_2\)): Molar mass = 32.00 g/mol \[ \text{Moles of O}_2 = \frac{48.0 \, \text{g}}{32.00 \, \text{g/mol}} = 1.500 \, \text{mol} \]

• Nitrogen (N\(_2\)): Molar mass = 28.02 g/mol \[ \text{Moles of N}_2 = \frac{14.0 \, \text{g}}{28.02 \, \text{g/mol}} = 0.4993 \, \text{mol} \]

• Disulfur Trioxide (S\(_2\)O\(_3\)): Molar mass = 112.13 g/mol \[ \text{Moles of S}_2\text{O}_3 = \frac{64.0 \, \text{g}}{112.13 \, \text{g/mol}} = 0.5708 \, \text{mol} \]

• Hydrogen (H\(_2\)): Molar mass = 2.016 g/mol \[ \text{Moles of H}_2 = \frac{9.0 \, \text{g}}{2.016 \, \text{g/mol}} = 4.4643 \, \text{mol} \]

Sum the moles of all gases to find the total moles in the mixture.

\[ \text{Total moles} = 0.4997 + 1.500 + 0.4993 + 0.5708 + 4.4643 = 7.5341 \, \text{mol} \]

The partial pressure of each gas is given by Dalton's Law of Partial Pressures:

\[ P_i = \left(\frac{\text{moles of } i}{\text{total moles}}\right) \times P_{\text{total}} \]

Convert the total pressure from kPa to atm: \(1 \, \text{atm} = 101.325 \, \text{kPa}\).

\[ P_{\text{total}} = \frac{800.0 \, \text{kPa}}{101.325 \, \text{kPa/atm}} = 7.8948 \, \text{atm} \]

• Partial Pressure of He: \[ P_{\text{He}} = \left(\frac{0.4997}{7.5341}\right) \times 7.8948 = 0.5235 \, \text{atm} \]

• Partial Pressure of O\(_2\): \[ P_{\text{O}_2} = \left(\frac{1.500}{7.5341}\right) \times 7.8948 = 1.5705 \, \text{atm} \]

• Partial Pressure of N\(_2\): \[ P_{\text{N}_2} = \left(\frac{0.4993}{7.5341}\right) \times 7.8948 = 0.5230 \, \text{atm} \]

• Partial Pressure of S\(_2\)O\(_3\): \[ P_{\text{S}_2\text{O}_3} = \left(\frac{0.5708}{7.5341}\right) \times 7.8948 = 0.5978 \, \text{atm} \]

• Partial Pressure of H\(_2\): \[ P_{\text{H}_2} = \left(\frac{4.4643}{7.5341}\right) \times 7.8948 = 4.6800 \, \text{atm} \]

Final Answer