Questions: A 100 L tank at 45 °C is filled with 110 g of sulfur tetrafluoride gas and 55.6 g of sulfur hexafluoride gas. You can assume both gases behave as ideal gases. Calculate the mole fraction and partial pressure of each gas, and the total pressure in the tank. Round each of your answers to 3 significant digits. Total pressure in tank: sulfur tetrafluoride: molar fraction partial pressure sulfur hexafluoride: molar fraction partial pressure □ atm □ atm □ atm □ atm □ atm

Solution Steps

First, we need to calculate the number of moles of sulfur tetrafluoride (SF\(_4\)) and sulfur hexafluoride (SF\(_6\)).

The molar mass of SF\(_4\) is: \[ M_{\text{SF}_4} = 32.06 + 4 \times 19.00 = 108.06 \, \text{g/mol} \]

The molar mass of SF\(_6\) is: \[ M_{\text{SF}_6} = 32.06 + 6 \times 19.00 = 146.06 \, \text{g/mol} \]

Now, calculate the number of moles of each gas: \[ n_{\text{SF}_4} = \frac{110 \, \text{g}}{108.06 \, \text{g/mol}} = 1.018 \, \text{mol} \] \[ n_{\text{SF}_6} = \frac{55.6 \, \text{g}}{146.06 \, \text{g/mol}} = 0.381 \, \text{mol} \]

The total number of moles in the tank is: \[ n_{\text{total}} = n_{\text{SF}_4} + n_{\text{SF}_6} = 1.018 + 0.381 = 1.399 \, \text{mol} \]

The mole fraction of SF\(_4\) is: \[ x_{\text{SF}_4} = \frac{n_{\text{SF}_4}}{n_{\text{total}}} = \frac{1.018}{1.399} = 0.728 \]

The mole fraction of SF\(_6\) is: \[ x_{\text{SF}_6} = \frac{n_{\text{SF}_6}}{n_{\text{total}}} = \frac{0.381}{1.399} = 0.272 \]

Using the ideal gas law \( PV = nRT \), where \( R = 0.0821 \, \text{L·atm/(mol·K)} \) and \( T = 45 \, ^\circ\text{C} = 318 \, \text{K} \):

\[ P = \frac{nRT}{V} = \frac{1.399 \times 0.0821 \times 318}{100} = 0.364 \, \text{atm} \]

The partial pressure of SF\(_4\) is: \[ P_{\text{SF}_4} = x_{\text{SF}_4} \times P_{\text{total}} = 0.728 \times 0.364 = 0.265 \, \text{atm} \]

The partial pressure of SF\(_6\) is: \[ P_{\text{SF}_6} = x_{\text{SF}_6} \times P_{\text{total}} = 0.272 \times 0.364 = 0.099 \, \text{atm} \]

Final Answer