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 \(100.0-\mathrm{mL}\) nickel container is filled with xenon and fluorine, giving partial pressures of \(1.24\) atm and \(10.10 \mathrm{~atm}\). 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 \(7.62 \mathrm{~atm}\). Assuming all of the xenon has \(\mathrm{re}\) acted, what is the formula of the product?

Step by step solution

01

Calculate the initial moles of Xe and F2

Using the ideal gas law equation (PV = nRT), we can find the initial moles of Xe and F2. The equation can be rearranged to solve for moles (n): \(n = \frac{PV}{RT}\) where P = pressure, V = volume, R = ideal gas constant (0.0821 L atm/(mol K)), and T = temperature in Kelvin. First, convert the given temperatures to Kelvin: Initial temperature (T1) = \(25^{\circ} \mathrm{C} + 273.15\) = 298.15 K Now, calculate the moles of Xe: \(n_{Xe} = \frac{P_{Xe}V}{RT_1} = \frac{1.24\, \mathrm{atm} \times 100.0\, \mathrm{mL}}{0.0821\, \mathrm{L\, atm /(mol\, K)} \times 298.15\, \mathrm{K}}\) And moles of F2: \(n_{F2} = \frac{P_{F2}V}{RT_1} = \frac{10.10\, \mathrm{atm} \times 100.0\, \mathrm{mL}}{0.0821\, \mathrm{L\, atm /(mol\, K)} \times 298.15\, \mathrm{K}}\)

02

Calculate the moles of F2 after reaction

Since the remaining F2 gas is transferred to another container of the same volume, we can find the moles of F2 after the reaction using the same ideal gas law equation with the final pressure (7.62 atm) and the initial temperature (T1) in Kelvin. \(n'_{F2} = \frac{P'_{F2}V}{RT_1} = \frac{7.62\, \mathrm{atm} \times 100.0\, \mathrm{mL}}{0.0821\, \mathrm{L\, atm /(mol\, K)} \times 298.15\, \mathrm{K}}\)

03

Calculate the moles of Xe and F2 in the product

Now that we have calculated the moles of F2 before and after the reaction, we can then find the number of F2 moles consumed in the reaction. Since all of the xenon has reacted, the moles of Xe in the reaction will be equal to the initial moles of Xe calculated in step 1. Moles of F2 consumed (n_F2_consumed) = moles of F2 before reaction - moles of F2 after reaction

04

Determine the formula of the xenon fluoride compound

To determine the formula of the xenon fluoride compound, we need to find the ratio of xenon to fluorine atoms in the compound. We can do this by dividing the moles of F2 consumed by the moles of Xe in the reaction. Ratio (F2:Xe) = (n_F2_consumed)/(n_Xe) Since the ratio of moles doesn't necessarily represent the ratio of atoms in the compound, we need to check whether the obtained ratio needs simplification or not. If the ratio does need simplification, divide the numbers in the ratio by their greatest common divisor (GCD). Once the simplest whole number ratio is determined, we can write the formula of the xenon fluoride compound as XeFx, where x is the simplified ratio (number of fluorine atoms) in the product.

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