Xenon and fluorine will react to form binary compounds when a mixture of these two gases is heated to \(400^{\circ} \mathrm{C}\) in a nickel reaction vessel. A 100.0 -mL nickel container is filled with xenon and fluorine, giving partial pressures of 1.24 atm and 10.10 atm, respectively, at a temperature of \(25^{\circ} \mathrm{C}\) . The reaction vessel is heated to $400^{\circ} \mathrm{C}$ to cause a reaction to occur and then cooled to a temperature at which \(\mathrm{F}_{2}\) is a gas and the xenon fluoride compound produced is a nonvolatile solid. The remaining \(\mathrm{F}_{2}\) gas is transferred to another \(100.0-\mathrm{mL}\) nickel container, where the pressure of \(\mathrm{F}_{2}\) at \(25^{\circ} \mathrm{C}\) is 7.62 \(\mathrm{atm}\) . Assuming all of the xenon has reacted, what is the formula of the product?

The ideal gas law formula is \(PV = nRT\), where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant (0.0821 L atm / K mol) and T is the temperature in Kelvin. Using this formula, we can calculate the initial moles of xenon and fluorine: For xenon: \(P=\) 1.24 atm, \(V=\) 100.0 mL = 0.1 L, \(R=\) 0.0821 L atm / K mol, \(T=\) (25 + 273.15) K = 298.15 K \(n_Xe = \frac{PV}{RT}\) \(n_Xe = \frac{(1.24 \,\mathrm{atm})(0.1 \,\mathrm{L})}{(0.0821\, \mathrm{L\, atm / K\, mol})(298.15\, \mathrm{K})}\) \(n_Xe \approx 0.00506 \,\mathrm{moles}\) For fluorine: \(P=\) 10.10 atm, \(V=\) 100.0 mL = 0.1 L, \(R=\) 0.0821 L atm / K mol, \(T=\) (25 + 273.15) K = 298.15 K \(n_F = \frac{PV}{RT}\) \(n_F = \frac{(10.10 \,\mathrm{atm})(0.1 \,\mathrm{L})}{(0.0821\, \mathrm{L\, atm / K\, mol})(298.15\, \mathrm{K})}\) \(n_F \approx 0.04083 \,\mathrm{moles}\)

We'll perform a similar calculation for the moles of remaining fluorine after the reaction: Remaining fluorine: \(P=\) 7.62 atm, \(V=\) 100.0 mL = 0.1 L, \(R=\) 0.0821 L atm / K mol, \(T=\) (25 + 273.15) K = 298.15 K \(n_F' = \frac{PV}{RT}\) \(n_F' = \frac{(7.62 \,\mathrm{atm})(0.1 \,\mathrm{L})}{(0.0821\, \mathrm{L\, atm / K\, mol})(298.15\, \mathrm{K})}\) \(n_F' \approx 0.03110 \,\mathrm{moles}\)

Subtract the moles of fluorine remaining after the reaction from the initial moles of fluorine to obtain the moles of fluorine that reacted with xenon: \(n_{F,reacted} = n_F - n_F'\) \(n_{F,reacted} = 0.04083 \,\mathrm{moles} - 0.03110 \,\mathrm{moles}\) \(n_{F,reacted} \approx 0.00973 \,\mathrm{moles}\)

Divide the moles of reacted fluorine by the moles of xenon to find the mole ratio between them in the compound: Mole ratio = \(\frac{n_{F,reacted}}{n_Xe}\) Mole ratio = \(\frac{0.00973 \,\mathrm{moles}}{0.00506 \,\mathrm{moles}}\) Mole ratio ≈ 1.922 Since the mole ratio must be an integer, we conclude that the mole ratio is 2.

For every mole of xenon, there are approximately 2 moles of fluorine. So the formula of the product should be: XeF₂ The formula of the xenon fluoride compound produced is XeF₂.