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 \(100.0\) -mL nickel container, where the pressure of \(\mathrm{F}_{2}\) at \(25^{\circ} \mathrm{C}\) is \(7.62\) atm. Assuming all of the xenon has reacted, what is the formula of the product?

Step by step solution

01

Calculate moles of Xenon and Fluorine initially

Use the ideal gas law, \(PV=nRT\), where \(P\) is pressure, \(V\) is volume, \(n\) is the moles of the gas, \(R\) is the ideal gas constant (0.08206 L.atm.K^(-1).mol^(-1)), and \(T\) is temperature in Kelvin. First, convert the temperature from Celsius to Kelvin. \(T = 25^{\circ} \mathrm{C} + 273.15 = 298.15 \mathrm{K}\) Now, calculate the moles of Xenon and Fluorine. For Xenon: \(n_{Xe} = \frac{P_{Xe}V}{RT} = \frac{1.24\,\text{atm}\times100.0\,\text{mL}}{0.08206\,\text{L}\cdot\text{atm}\cdot\text{K}^{-1}\cdot\text{mol}^{-1}\times298.15\,\text{K}} = 0.0506\,\text{moles}\) For Fluorine: \(n_{F_2} = \frac{P_{F_2}V}{RT} = \frac{10.10\,\text{atm}\times100.0\,\text{mL}}{0.08206\,\text{L}\cdot\text{atm}\cdot\text{K}^{-1}\cdot\text{mol}^{-1}\times298.15\,\text{K}} = 0.412\,\text{moles}\)

02

Calculate moles of Fluorine remaining

After the reaction, the pressure of Fluorine remaining is \(7.62\,\text{atm}\). Repeat the above calculations to find out the moles of remaining Fluorine. \(n_{F_2_{final}} = \frac{P_{F_2_{final}}V}{RT} = \frac{7.62\,\text{atm}\times100.0\,\text{mL}}{0.08206\,\text{L}\cdot\text{atm}\cdot\text{K}^{-1}\cdot\text{mol}^{-1}\times298.15\,\text{K}} = 0.311\,\text{moles}\)

03

Calculate moles of Fluorine reacted

Subtract the moles of Fluorine remaining from the initial moles of Fluorine. \(n_{F_2_{reacted}} = n_{F_2_{initial}} - n_{F_2_{final}} = 0.412\,\text{moles} - 0.311\,\text{moles} = 0.101\,\text{moles}\)

04

Calculate stoichiometry

Determine the stoichiometry of the reaction (the ratio of moles of Xenon to moles of Fluorine). \(\frac{n_{Xe}}{n_{F_2_{reacted}}} = \frac{0.0506\,\text{moles}}{0.101\,\text{moles}} \approx \frac{1}{2}\) Thus, the stoichiometry of the reaction is approximately 1:2 (Xe:F2).

05

Write the formula of the product

The product is a binary compound formed by the reaction of Xenon and Fluorine. Since the stoichiometry of the reaction is 1:2, the formula of the product is \(\text{XeF}_2\).

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