A 20.0-L nickel container was charged with \(0.500\) atm of xenon gas and \(1.50\) 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 and fluorine is \(Xe + 2F_2 \rightarrow XeF_4\). Using the ideal gas law equation and the given conditions, we can find that the number of moles of xenon is \(n_{Xe} = \frac{0.5 atm \times 20.0 L}{0.0821 \frac{L \cdot atm}{mol \cdot K} \times 673.15 K}\) and the number of moles of fluorine is \(n_{F_2} = \frac{1.5 atm \times 20.0 L}{0.0821 \frac{L \cdot atm}{mol \cdot K} \times 673.15 K }\). Comparing their moles with their stoichiometric coefficients, we can determine the limiting reactant, and calculate the moles of xenon tetrafluoride produced. Finally, multiply the moles of xenon tetrafluoride by its molar mass (207.29 g/mol) to get the mass of xenon tetrafluoride produced, assuming 100% yield: \(Mass\ XeF_4 = \ moles\ XeF_4 \times Molar\ Mass\ XeF_4\).