Chapter 9, Problem 47A

Interpretation Introduction

Interpretation: The information regarding the inert gases and formation of stable compounds by xenon and its reaction with fluorine using nickel catalyst is given. The theoretical mass of xenon tetrafluoride formed after reaction of xenon with fluorine and percent yield of xenon terafluoride is to be stated.

Concept introduction: The reactant that gets completely utilized in the reaction is defined as limiting reagent.

The number of moles is calculated as shown below:

  $\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ (1)

The formula to calculate percent yield is shown below:

  $\text{Percent yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100\%$ . (2)

Answer to Problem 47A

The theoretical mass of xenon tetrafluoride is $205.25\text{g}$ . The percent yield is $70.65\%$ .

Explanation of Solution

Given information : The given mass of xenon is $130.0\text{g}$ . The given mass of fluorine is $100.0\text{g}$ . The amount of xenon tetrafluoride isolated is $145\text{g}$ .

The molar mass of xenon is $131.293\text{g/mol}$ . The given mass is $130.0\text{g}$ . Substitute these values in above equation to calculate the number of moles.

  $\begin{array}{c}\text{Number of moles}=\frac{\text{130}\text{.0}\text{g}}{\text{131}\text{.293}\text{g/mol}}\\ =0.9902\text{mol}\end{array}$

The molar mass of fluorine is $37.9968\text{g/mol}$ . The given mass is $100.0\text{g}$ . Substitute these values in above equation to calculate the number of moles.

  $\begin{array}{c}\text{Number of moles}=\frac{\text{100}\text{.0}\text{g}}{37.9968\text{g/mol}}\\ =2.632\text{mol}\end{array}$

The given chemical equation is shown below:

  $\text{Xe}\left(g\right)+2{\text{F}}_2\left(g\right)\to {\text{XeF}}_{\text{4}}\left(s\right)$

The above equation shows that one mole of xenon reacts with two moles of fluorine gas. The limiting reagent is determined as shown below:

  $\begin{array}{c}\text{2 mol fluorine}=\text{1 mol xenon}\\ \text{2}\text{.632 mol fluorine}=\frac{\text{1}}{2}\times \text{2}\text{.632 mol xenon}\\ =1.\text{136 mol xenon}\end{array}$

Thus, xenon acts as a limiting reagent.

The number of moles of xenon tetrafluoride generated from given amount of xenon is calculated as shown below:

  $\begin{array}{c}\text{1 mol Xe}={\text{1 mol XeF}}_{\text{4}}\\ \text{0}\text{.9902 mol Xe}=0.9902{\text{ mol XeF}}_{\text{4}}\\ =0.9902{\text{ mol XeF}}_{\text{4}}\end{array}$

The molar mass of xenon tetrafluoride is $207.2836\text{g/mol}$ . Substitute the values in equation (1) to calculate the mass of xenon tetrafluoride as shown below:

  $\begin{array}{c}\text{0}\text{.9902}\text{mol}=\frac{\text{Given mass}}{\text{207}\text{.2836}\text{g/mol}}\\ \text{Given mass}=205.25\text{g}\end{array}$

Substitute the values in equation (2) to calculate the percent yield as shown below:

  $\begin{array}{c}\text{Percent yield}=\frac{\text{145}\text{g}}{\text{205}\text{.25}\text{g}}\times 100\%\\ =70.65\%\end{array}$

Conclusion

The theoretical mass of xenon tetrafluoride is $205.25\text{g}$ . The percent yield is $70.65\%$ .