Questions: You have 2.2 mol Xe and 3.0 mol F2, but when you carry out the reaction you end up with only 0.25 mol XeF4. What is the percent yield of this experiment? Xe(g) + 2 F2(g) -> XeF4(g)

You have 2.2 mol Xe and 3.0 mol F2, but when you carry out the reaction you end up with only 0.25 mol XeF4. What is the percent yield of this experiment? Xe(g) + 2 F2(g) -> XeF4(g)

Transcript text: You have 2.2 mol Xe and $3.0 \mathrm{~mol} \mathrm{~F}_{2}$, but when you carry out the reaction you end up with only $0.25 \mathrm{~mol} \mathrm{XeF}_{4}$. What is the percent yield of this experiment? $\mathrm{Xe}(\mathrm{g})+2 \mathrm{~F}_{2}(\mathrm{~g}) \rightarrow \mathrm{XeF}_{4}(\mathrm{~g})$

Solution

Solution Steps

Step 1: Determine the Limiting Reactant

The balanced chemical equation is:

\[ \text{Xe}(\text{g}) + 2 \text{F}_2(\text{g}) \rightarrow \text{XeF}_4(\text{g}) \]

From the equation, 1 mole of Xe reacts with 2 moles of $ \text{F}_2 $ to produce 1 mole of $ \text{XeF}_4 $.

Moles of Xe available: 2.2 mol
Moles of $ \text{F}_2 $ available: 3.0 mol

Calculate the moles of $ \text{F}_2 $ required to react with 2.2 mol of Xe:

\[ 2.2 \, \text{mol Xe} \times \frac{2 \, \text{mol} \, \text{F}_2}{1 \, \text{mol Xe}} = 4.4 \, \text{mol} \, \text{F}_2 \]

Since only 3.0 mol of $ \text{F}_2 $ is available, $ \text{F}_2 $ is the limiting reactant.

Step 2: Calculate Theoretical Yield

Using the limiting reactant ($ \text{F}_2 $), calculate the theoretical yield of $ \text{XeF}_4 $:

\[ 3.0 \, \text{mol} \, \text{F}_2 \times \frac{1 \, \text{mol} \, \text{XeF}_4}{2 \, \text{mol} \, \text{F}_2} = 1.5 \, \text{mol} \, \text{XeF}_4 \]

Step 3: Calculate Percent Yield

The actual yield of $ \text{XeF}_4 $ is given as 0.25 mol. The percent yield is calculated as follows:

\[ \text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100\% \]

\[ \text{Percent Yield} = \left( \frac{0.25 \, \text{mol}}{1.5 \, \text{mol}} \right) \times 100\% = 16.67\% \]