You have 2.2 mol Xe and 2.0 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)

The correct answer is 25 %

Explanation:

According to the chemical reaction:

Xe(g) + 2 F₂ (g) → XeF₄ (g)

1 mol of Xe(g) reacts with 2 mol of F₂(g), so the stoichiometric ratio os reactants is 2 mol F₂/mol Xe.

We have 2.2 mol Xe and 2.0 mol F₂, so the ratio is:

2.0 mol F₂/2.2 mol Xe = 0.909 mol F₂/mol Xe

The molar ratio of reactant we have is lower than the required, so F₂ is the limiting reactant.

By using the limiting reactant, we calculate the theoretical amount of product (XeF₄). For this, we know that 1 mol of XeF₄ is formed from 2 mol of F₂ (1 mol XeF₄/ 2 mol F₂), and we have 2.0 mol F₂:

2.0 mol F₂ x (1 mol XeF₄/ 2 mol F₂)= 1 mol XeF₄

If we only obtained 0.25 mol XeF₄, the percent yield of the experiment is:

Yield = experimental amount/theoretical amount x 100%

        = 0.25 mol XeF₄/ 1 mol XeF₄ x 100% = 25 %