Question

A \(20.0\) -\(\mathrm{L}\) 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 produced, assuming 100% yield, is \(60.23\ g\).

Step-by-step solution

Write the balanced chemical equation

The reaction between xenon and fluorine produces xenon tetrafluoride. The balanced chemical equation for this reaction is: \[Xe + 2 F_2 \rightarrow XeF_4\]

Calculate the initial moles of reactants

We are given the initial pressures and volume of the container. To find the moles of xenon and fluorine, 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 gas constant (which is \(0.0821\frac{L\cdot atm}{mol\cdot K}\) ), and T is the temperature in Kelvin. First, we have to convert the provided temperature from Celsius to Kelvin: \[T = 400^{\circ}C + 273.15 = 673.15 K\] Now, we can calculate the initial moles of xenon and fluorine: Moles of xenon (nXe): \[0.859\ atm \cdot 20.0\ L = nXe \cdot 0.0821\frac{L\cdot atm}{mol\cdot K} \cdot 673.15 K\] \[nXe = \frac{0.859 \cdot 20.0}{0.0821 \cdot 673.15} = 0.2127\ mol\] Moles of fluorine (nF2): \[1.37\ atm \cdot 20.0\ L = nF2 \cdot 0.0821\frac{L\cdot atm}{mol\cdot K} \cdot 673.15 K\] \[nF2 = \frac{1.37 \cdot 20.0}{0.0821 \cdot 673.15} = 0.4268\ mol\]

Apply stoichiometry and identify limiting reactant

Now, let's determine the limiting reactant and calculate the moles of xenon tetrafluoride produced. From the balanced equation, we can see that 1 mole of xenon reacts with 2 moles of fluorine to produce 1 mole of xenon tetrafluoride. The mole ratio is 1:2:1. Now we will use the mole ratios to find out which reactant runs out first (the limiting reactant) and how many moles of xenon tetrafluoride are produced: Moles of XeF4 produced from Xe: \[\frac{0.2127\ mol\ Xe}{1\ mol\ Xe} = 0.2127\ mol\ XeF_4\] Moles of XeF4 produced from F2: \[\frac{0.4268\ mol\ F2}{2\ mol\ F2} = 0.2134\ mol\ XeF_4\] As we can see, xenon will produce fewer moles of xenon tetrafluoride, so xenon is the limiting reactant and 0.2127 moles of XeF4 will be produced.

Calculate mass of xenon tetrafluoride produced

Now that we have the moles of xenon tetrafluoride produced, we can find the mass by using the molar mass of XeF4: \[molar\ mass\ of\ XeF_4 = 131.29\ g/mol\ Xe + (4 \times 18.998\ g/mol\ F) = 283.29 g/mol\] Finally, we calculate the mass of xenon tetrafluoride: \[mass\ of\ XeF_4 = 0.2127\ mol\ XeF_4 \times 283.29\frac{g}{mol} = 60.23\ g\] The mass of xenon tetrafluoride produced, assuming 100% yield, is 60.23 grams.

Key concepts

Ideal Gas Law

Understanding the Ideal Gas Law is crucial for solving a variety of problems in chemistry, particularly when dealing with gases under certain conditions. Typically represented by the equation \(PV = nRT\), each variable within this equation is essential for accurately calculating the behavior of an ideal gas.

\(P\) stands for pressure, measured in atmospheres (atm), while \(V\) is the volume in liters (L). The \(n\) represents the amount of gas in moles, and \(T\) is the temperature in Kelvin (K). Lastly, \(R\) is the universal gas constant, valued at \(0.0821\frac{L\cdot atm}{mol\cdot K}\).

When solving problems like the one in our exercise, we use the Ideal Gas Law to find the initial moles of gas present in the container before reacting. This is the groundwork for further stoichiometric calculations.

Balancing Chemical Equations

To delve into any chemical reaction, such as the formation of xenon tetrafluoride from xenon and fluorine, the first step is to balance the chemical equation. Balancing equations ensures that the number of atoms for each element is the same on both the reactant and product sides according to the Law of Conservation of Mass.

The general process involves adjusting the coefficients (the numbers before each species) to get the same number of each type of atom on both sides. For instance, in our example \[Xe + 2 F_2 \rightarrow XeF_4\], the '2' before \(F_2\) indicates that two molecules of fluorine are required to react with one atom of xenon to produce one formula unit of xenon tetrafluoride, abiding by the stoichiometry of the equation.

Limiting Reactant

The limiting reactant in a chemical reaction is the substance that is entirely consumed when the chemical reaction is complete. This reactant determines the maximum amount of product that can be formed and thus is critical for stoichiometric calculations.

To identify the limiting reactant, compare the mole ratio of the reactants used to the mole ratio from the balanced equation. The reactant that provides the lesser amount of product (as dictated by the balanced equation) is the limiting reactant. In our exercise, we compared the moles of xenon (\(nXe\)) and fluorine (\(nF2\)) that could form xenon tetrafluoride (\(XeF4\)). Despite having more moles of fluorine, xenon was the limiting reactant due to its smaller stoichiometric coefficient in the balanced equation.

Molar Mass Calculation

Calculating the molar mass is a vital step in converting between the mass of a substance and the number of moles. The molar mass is the weight of one mole of a substance, typically expressed in grams per mole (g/mol).

To find the molar mass, sum the atomic masses of all atoms in the molecule. For xenon tetrafluoride, you add the mass of one xenon atom (\(131.29 g/mol\)) to the mass of four fluoride atoms (\(4 \times 18.998 g/mol\)), giving us \(283.29 g/mol\).

Once you know the moles of the limiting reactant and the molar mass of the product, you can calculate the theoretical yield of the product. In this particular exercise, multiplying the moles of xenon tetrafluoride by its molar mass provides the mass of the compound that would be produced in a 100% yield reaction.