A 20.0-L nickel container was charged with \(0.859\) atm of xenon gas and \(1.37\) atm of fluorine gas at \(400^{\circ} \mathrm{C}\). The xenon and fluorine react to form xenon tetrafluoride. What mass of xenon tetrafluoride can be produced assuming \(100 \%\) yield?

The balanced chemical equation for the reaction between xenon gas (Xe) and fluorine gas (F2) to form xenon tetrafluoride (XeF4) is: \[Xe_{(g)} + 2F_{2(g)} \rightarrow XeF_{4(s)}\] This equation tells us that one mole of xenon gas reacts with two moles of fluorine gas to produce one mole of xenon tetrafluoride.

We are given the partial pressure of each gas. To find the moles of each gas, we can use the ideal gas law equation: \[PV = nRT\] Where: - P is the pressure - V is the volume - n is the number of moles - R is the ideal gas constant (which is equal to 0.0821 L atm / K mol) - T is the temperature in Kelvin First, let's convert the given temperature from Celsius to Kelvin: \[T_K = T_C + 273 = 400 + 273 = 673 K\] Now let's calculate the moles for each gas: - Moles of xenon (n_Xe): \[0.859 \text{ atm} \times 20.0 L = n_Xe(0.0821 \frac{\text{L atm}}{\text{K mol}})(673 \text{K})\] \[n_Xe = \frac{0.859 \times 20.0}{0.0821 \times 673} = 0.2536 \text{ mol}\] - Moles of fluorine (n_F2): \[1.37 \text{ atm} \times 20.0 L = n_{F2}(0.0821 \frac{\text{L atm}}{\text{K mol}})(673 \text{K})\] \[n_{F2} = \frac{1.37 \times 20.0}{0.0821 \times 673} = 0.4037 \text{ mol}\]

Based on the stoichiometry of the balanced chemical equation, we need 2 moles of F2 for every 1 mole of Xe. So, we can calculate the amount of fluorine gas needed for the available xenon gas: \[0.2536 \text{ mol Xe} \times \frac{2\text{ mol F}_2}{1 \text{ mol Xe}} = 0.5072\text{ mol F}_2\] We find that 0.5072 moles of F2 are needed, but we only have 0.4037 moles of F2. Thus, fluorine is the limiting reactant.

Since fluorine is the limiting reactant, let's calculate the number of moles of xenon tetrafluoride produced from the reaction: \[0.4037 \text{ mol F}_2 \times \frac{1\text{ mol XeF}_4}{2\text{ mol F}_2} = 0.20185\text{ mol }\text{XeF}_4\] Now we can calculate the mass of xenon tetrafluoride produced by using its molar mass (Molecular weight of Xe = 131.29 g/mol, Molecular weight of F = 19.00 g/mol): \[0.20185 \text{ mol }\text{XeF}_4 \times \frac{131.29 + 4 \times 19.00 \text{ g}}{\text{1 mol }} = 57.46 \text{ g}\] Therefore, assuming 100% yield, the mass of xenon tetrafluoride that can be produced is 57.46 g.