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

$$\ce{2 P + 3 Cl2 -&gt; 2 PCl3}\tag{2}$$

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:

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

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

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

Since $0.33 &lt; 0.39$, 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:

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

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.