Answer:
Percentage yield = 23.8%
Step-by-step 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%