Question

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?

Short answer

The mass of xenon tetrafluoride that can be produced, assuming 100% yield, is 57.46 g.

Step-by-step solution

Write the balanced chemical equation for the reaction

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.

Calculate the moles of the reactants

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): \[1.37 \text{ atm} \times 20.0 L = n_{F2}(0.0821 \frac{\text{L atm}}{\text{K mol}})(673 \text{K})\] \[n_{F2} = \frac{1.37 \times 20.0}{0.0821 \times 673} = 0.4037 \text{ mol}\]

Identify the limiting reactant

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.

Calculate the mass of xenon tetrafluoride produced

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.

Key concepts

Ideal Gas Law

The ideal gas law is a fundamental principle in chemistry that describes the behavior of gases. It can be expressed as \(PV = nRT\), where \(P\) represents 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 temperature in Kelvin. This equation allows us to calculate the amount of gas in moles when pressure, volume, and temperature are known.

In this exercise, to determine the number of moles of xenon and fluorine gases, the ideal gas law was used after converting the temperature from Celsius to Kelvin. By rearranging the equation to \(n = \frac{PV}{RT}\), we can find the moles of each gas in the container. This is essential to understanding the stoichiometry of the reaction between xenon and fluorine to produce xenon tetrafluoride.

Limiting Reactant

The limiting reactant in a chemical reaction is the substance that is entirely consumed, determining the maximum amount of product formed. Identifying the limiting reactant involves comparing the mole ratio of the reactants used in the reaction to those required by the balanced chemical equation.

In the problem, the reaction requires 2 moles of fluorine gas for each mole of xenon gas. From the calculated moles of xenon \(n_{Xe} = 0.2536\) mol and fluorine \(n_{F_2} = 0.4037\) mol, we find that we don't have enough fluorine to react with all available xenon. Calculating the required 0.5072 moles of fluorine to react fully with xenon highlights that fluorine is the limiting reactant, stopping any further production of xenon tetrafluoride.

Balanced Chemical Equation

A balanced chemical equation is essential to understanding stoichiometry in chemical reactions. It ensures the law of conservation of mass is met, meaning the number of atoms of each element is the same on both sides of the reaction. In this exercise, the balanced equation is \( \text{Xe}_{(g)} + 2\text{F}_{2(g)} \rightarrow \text{XeF}_{4(s)} \).

This equation communicates that one mole of xenon reacts with two moles of fluorine gas to produce one mole of xenon tetrafluoride. Knowing this ratio is critical for identifying the limiting reactant and calculating the mass of the product that can be formed, determining how much product will be produced in any given reaction.

Molar Mass Calculation

Calculating molar mass is crucial in converting moles to grams, which is essential for determining the mass of products in a reaction. Molar mass is found by adding the atomic masses of the elements in a compound. For example, in xenon tetrafluoride (\( \text{XeF}_4 \)), the molar mass is calculated by adding the atomic mass of xenon (131.29 g/mol) to four times the atomic mass of fluorine (19.00 g/mol).

In the solution, the molar mass of \( \text{XeF}_4 \) is used to convert the calculated 0.20185 moles of \( \text{XeF}_4 \) into grams, resulting in a mass of 57.46 grams. This conversion is critical for quantifying the amount of a substance involved in or resulting from a chemical reaction.