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) - XeF. (g)

25%

Explanation:

In this case, let's write the reaction again:

Xe(g) + 2F₂(g) > XeF₄(g)

Now, we know that we got 0.25 mol of the product. We want to know the percent yield. To get this value, we need to calculate the theorical yield.

We have 2.2 mol of Xe and 2 mol of F₂, we need to know the limiting reactant.

According to the reaction, the mole ratio:

1 moles Xe > 2 moles F₂

According to this, and the given data, we have 2.2 moles of Xe, so:

1 moles Xe > 2 moles F₂

2.2 moles Xe > X

X = 2.2 * 2 / 1 = 4.4 moles of F₂

However we only have 2 moles of F₂, therefore, the limiting reactant is F₂ while the excess is the Xe.

So the actual moles reacting here would be 2 moles of F₂, and this will produce X moles of XeF₄:

2 mole of F₂ > 1 mole XeF₄

The moles of XeF₄ would be 1 mole theorical.

If we got 0.25 then the percent yield would have to be:

% = 0.25 / 1 * 100

% = 25%

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