Question

Although they were formerly called the inert gases, at least the heavier elements of Group 8 do form relatively stable compounds. For example, xenon combines directly with elemental fluorine at elevated temperatures in the presence of a nickel catalyst. $$ \mathrm{Xe}(g)+2 \mathrm{~F}_{2}(g) \rightarrow \mathrm{XeF}_{4}(s) $$ What is the theoretical mass of xenon tetrafluoride that should form when \(130 .\) g of xenon is reacted with 100.8 of \(\mathrm{F}_{2} ?\) What is the percent yield if only \(145 \mathrm{~g}\) of \(\mathrm{XeF}_{4}\) is actually isolated?

Short answer

The theoretical mass of xenon tetrafluoride (\(\mathrm{XeF}_{4}\)) that should form when \(130g\) of xenon (\(\mathrm{Xe}\)) is reacted with \(100.8g\) of fluorine (\(\mathrm{F}_{2}\)) is approximately \(205.14g\). The percent yield of \(\mathrm{XeF}_{4}\) when only \(145g\) is actually isolated is approximately \(70.7\%\).

Step-by-step solution

Calculate the moles of reactants

To begin, we need to convert the masses of the reactants into moles. The molar mass of \(\mathrm{Xe}\) is approximately \(131.29 \frac{g}{mol}\) and the molar mass of \(\mathrm{F}_{2}\) is approximately \(38 \frac{g}{mol}\). Moles of \(\mathrm{Xe} = \frac{130g}{131.29 \frac{g}{mol}} \approx 0.9903\ \mathrm{mol\ of\ Xe}\) Moles of \(\mathrm{F}_{2} = \frac{100.8g}{38 \frac{g}{mol}} \approx 2.6526\ \mathrm{mol\ of\ F_2}\)

Determine the limiting reactant

Now we can determine the limiting reactant by comparing the ratio of moles to the stoichiometry of the balanced equation. The balanced equation has a 1:2 stoichiometry, which means 1 mole of xenon will react with 2 moles of fluorine. Let's divide the moles of each reactant by their respective stoichiometric coefficients: For \(\mathrm{Xe}\): \(\frac{0.9903 \ mol \ of \ Xe}{1} = 0.9903\) For \(\mathrm{F}_{2}\): \(\frac{2.6526 \ mol \ of \ F_2}{2} = 1.3263\) The lower value, \(0.9903\), belongs to \(\mathrm{Xe}\), so it is the limiting reactant.

Calculate the theoretical mass of XeF4

Now, we will use the stoichiometry of the balanced equation to find the theoretical mass of \(\mathrm{XeF}_{4}\). The molar mass of \(\mathrm{XeF}_{4}\) is approximately \(207.29 \frac{g}{mol}\). Moles of \(\mathrm{XeF}_{4} = 0.9903 \ mol\ of\ Xe\cdot \frac{1 \ mol\ of\ XeF_{4}}{1 \ mol\ of\ Xe} = 0.9903 \ mol\ of\ XeF_{4}\) Theoretical mass of \(\mathrm{XeF}_{4} = 0.9903\ \mathrm{mol\ of\ XeF_{4}} \cdot 207.29 \frac{g}{mol} \approx 205.14 g\)

Calculate the percent yield

Now we can calculate the percent yield using the actual mass of \(\mathrm{XeF}_{4}\) produced and the theoretical mass calculated in step 3. Percent Yield = \(\frac{Actual \ Mass}{Theoretical \ Mass} \times 100\) Percent Yield = \(\frac{145g}{205.14g} \times 100 \approx 70.7\%\) So, the percent yield of \(\mathrm{XeF}_{4}\) is approximately \(70.7\%\).

Key concepts

Stoichiometry

Stoichiometry is the branch of chemistry that deals with the quantities of substances that interact in chemical reactions. It is based on the principle of the conservation of mass, where the total mass of reactants equals the total mass of products in a chemical equation. To understand stoichiometry, one must be familiar with the chemical equation and how it represents the ratio of reactants to products. For instance, in the reaction of xenon with fluorine to form xenon tetrafluoride (\(\mathrm{Xe}(g) + 2 \mathrm{F_2}(g) \rightarrow \mathrm{XeF_4}(s)\)), the coefficients indicate the stoichiometric ratio: one mole of xenon reacts with two moles of fluorine to produce one mole of xenon tetrafluoride.

Calculations in stoichiometry often involve converting mass to moles, using the molar mass of each substance. By doing this, chemists can predict the amounts of products formed from given reactants, which is known as the theoretical yield. In our exercise, the theoretical mass of xenon tetrafluoride is calculated using the moles of the limiting reactant, xenon, based on the balanced equation provided.

Additionally, stoichiometry allows for the determination of the percent yield, which compares the amount of product actually obtained (actual yield) to the theoretical yield. The percent yield provides insight into the efficiency of the reaction and indicates the presence of any practical limitations that might reduce the yield.

Limiting Reactant

In a chemical reaction, the limiting reactant (or reagent) is the substance that is completely consumed first, thereby determining the maximum amount of product that can be formed. This concept is critical for carrying out stoichiometric calculations, as it allows chemists to calculate the exact amount of product that can be expected from a reaction. The limiting reactant sets a 'limit' on the reaction because once it is used up, the reaction cannot proceed further, regardless of the quantities of other reactants present.

In our example, xenon (\(\mathrm{Xe}\)) is the limiting reactant because it has the smaller mole ratio compared to fluorine (\(\mathrm{F_2}\)) when we look at the stoichiometry of the balanced chemical equation. To identify the limiting reactant, one must calculate the number of moles of each reactant and then divide by the respective coefficients in the balanced equation. The reactant yielding the lowest value is the one that will run out first, thus limiting how much product can be produced. This concept is crucial for predicting theoretical yields and ensuring that reactions are carried out efficiently with minimal waste.

Mole Concept

The mole concept is a fundamental principle in chemistry that serves as a bridge between the microscopic world of atoms and molecules and the macroscopic world of grams and liters. A mole represents a specific number of particles, equal to Avogadro's number (\(6.022 \times 10^{23}\) particles). It is used to count atoms, ions, or molecules in a given substance and is analogous to using 'dozen' to count eggs.

To perform stoichiometric calculations, chemists use the mole concept to convert masses of substances into moles, based on their molar masses (the mass of one mole of a substance). This conversion is what enables them to use the balanced equation to relate the amounts of various reactants and products to one another. In the example with xenon and fluorine, the moles of xenon and fluorine were calculated from their given masses and molar masses. These calculations led to finding moles of \(\mathrm{XeF_4}\) which, when multiplied by its molar mass, yielded the theoretical mass of the product. Understanding the mole concept is essential for understanding stoichiometry as it provides a consistent method to quantify chemical reactions.