You have 2.2 mol Xe and 2.1 mol F2, but when you carry out the reaction you end up with only 0.25 mol XeFg. What is the percent
yield of this experiment?
Xe(g) + 2 F2 (g) XeF, (g)

Percentage yield = 23.8%

Explanation:

We'll begin by determining the limiting reactant. This can be obtained as follow:

Xe(g) + 2F₂(g) —> XeF₄

From the balanced equation above,

1 mole of Xe reacted with 2 moles of F₂.

Therefore, 2.2 moles of Xe will react with = 2.2 × 2 = 4.4 moles of F₂.

From the above calculation, we can see that a higher amount (i.e 4.4 moles) of F₂ than what was given (i.e 2.1 moles) is needed to react completely with 2.2 moles Xe.

Therefore, F₂ is the limiting reactant and Xe is the excess reactant.

Next, we shall determine the theoretical yield of XeF₄.

This can be obtained by using the limiting reactant as follow:

From the balanced equation above,

2 moles of F₂ reacted to produce 1 mole of XeF₄.

Therefore, 2.1 moles of F₂ will react to produce = (2.1 × 1)/2 = 1.05 moles of XeF₄.

Thus, the theoretical yield of XeF₄ is 1.05 moles.

Finally, we shall determine the percentage yield. This can be obtained as follow:

Actual yield of XeF₄ = 0.25 mole

Theoretical yield of XeF₄ = 1.05 moles

Percentage yield =?

Percentage yield = Actual yield / Theoretical yield × 100

Percentage yield = 0.25 / 1.05 × 100

Percentage yield = 0.238 × 100

Percentage yield = 23.8%

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