# How to calculate the percentage yield for a reaction with excess reagent?

A chemist mixed $$12~\mathrm{g}$$ of phosphorus with $$35.5\ \mathrm{g}$$ of chlorine gas to synthesize phosphorus(III) chloride (phosphorus trichloride). The yield was $$42.4~\mathrm{g}$$ of $$\ce{PCl3}$$. The equation is $$\ce{2P + 3Cl2 -> 2PCl3}$$ Calculate the percentage yield.

I think that the amount of phosphorus the chemist used is in excess. I figured that only 1/3 of the amount of phosphorus the chemist used reacted to produce phosphorus trichloride. But the question didn't mention anything about products in excess that is why I am not so sure of my answer, which is $$92.51$$.

First I found the amount of substance of phosphorus used which is $$12/31$$. Then I calculated the amount of substance of chlorine gas used (and it's $$0.5$$). But $$12/31\ \mathrm{mol}$$ of $$\ce{P}$$ needs $$1.5 \times (12/31)\ \mathrm{mol}$$ of chlorine gas, so $$\ce{P}$$ is in excess. The amount of substance of $$\ce{P}$$ that will react will be $$0.5/1.5 = 1/3$$ Now $$1/3\ \mathrm{mol}$$ of $$\ce{P}$$ produces $$1/3\ \mathrm{mol}$$ of $$\ce{PCl3}$$ and the molecular mass of $$\ce{PCl3}$$ is $$137.5$$ so the mass that is supposed to be produced is $$137.5 \times 1/3$$. Then I divided $$42.4$$ by $$(137.5 \times 1/3)$$ and multiplied by $$100$$.

The yield is typically calculated according to equation $(1)$.

$$\text{yield} = \frac{\text{actual amount of substance obtained}}{\text{theoretical amount of substance to be obtained}}\tag{1}$$

(Whether it is given in percent or as a fractional number is not important, but percentages are the usual value.) So for any reaction, you need to figure out:

• the reaction equation
• the amount of each reactant substance involved
• the amount of desired product theoretically possible by stoichiometric coefficients
• the actual amount of desired product.

The first is rather easy, especially since the equation $(2)$ has been given previously.

$$\ce{2 P + 3 Cl2 -> 2 PCl3}\tag{2}$$

Thus, for every mole of phosphorus, we expect one mole of product and for each mole of chlorine gas, we expect two-thirds of a mole of product. The amounts of the reactants are easily calculated using molar masses:

\begin{align}M &= \frac mn\tag{3}\\ n (\ce{P}) &= \frac{12~\mathrm{g}}{30.97~\mathrm{g/mol}}\\ &= 0.39~\mathrm{mol}\tag{4}\\[1em] n(\ce{Cl2}) &= \frac{35.5~\mathrm{g}}{70.9~\mathrm{g/mol}}\\ &= 0.50~\mathrm{mol}\tag{5}\end{align}

For each of these two we would need to check the amount of product we would expect using the stoichiometric coefficients.

\begin{align}n_\text{theor}(\ce{PCl3}) &= \frac 23 n(\ce{Cl2})\\ &= 0.33~\mathrm{mol}\tag{6}\\[1em] n_\text{theor}(\ce{PCl3}) &= n(\ce{P})\\ &= 0.39~\mathrm{mol}\tag{7}\end{align}

Since $0.33 < 0.39$, chlorine is the limiting reagent and our theoretically possible yield is calculated from the amount of chlorine added to the reaction. Therefore, we now know our denominator which is $0.33~\mathrm{mol}$. What is the numerator? Again, it is easily calculated:

\begin{align}n_\text{exp}(\ce{PCl3}) &= \frac{42.4~\mathrm{g}}{137.32~\mathrm{g/mol}}\\ &= 0.31~\mathrm{mol}\end{align}\tag{8}

And now all we need to do is calculate the yield:

$$\frac{0.31~\mathrm{mol}}{0.33~\mathrm{mol}} = 0.93 = 93~\%\tag{9}$$

This is consistent with your result.

I'd of done this using a different approach.

Frankly I'd first try a shortcut of sorts. Such chemistry problems usually consume all of one of the reactants.

Knowing that 42.4 g of $\ce{PCl3}$ were produced

g P = $\dfrac{30.97}{137.3} \cdot 42.4 = 9.56$ grams

the rest must be Cl so

g Cl = 42.4 - 9.6 = 32.8 grams

Since neither of these amounts is equal to the mass of the starting reagents I need to look at the moles of the reactants to determine which is the limiting reagent.

moles P = $\dfrac{12}{30.97} = 0.387$

moles Cl atoms = $\dfrac{35.5}{35.5} = 1$

But every mole of $\ce{PCl3}$ needs 3 moles of Cl atoms. So 35.5 grams of Cl can make at most 0.333 moles of $\ce{PCl3}$.

So theoretically the yield should have been
$0.333 \cdot 137.3 = 45.8$ grams

% Yield = $\dfrac{42.4}{45.8}\cdot 100\% = 92.6\%$