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


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