Question

The percent yield for the reaction, $PC{l}_3+C{l}_2⟶PC{l}_5$ is $83.2\%$. What mass of $PC{l}_5$ would be expected from the reaction of $56.7$ g of $PC{l}_3$ with excess $C{l}_2$?
$(P=31,Cl=35.5)$ (If answer is $x$, then write answer to the nearest integer).

Answer 1

The molar masses of $PC{l}_3$ and $PC{l}_5$ are $137.33$ g/mol and $208.24$ g/mol respectively.
The expected mass of $PC{l}_5$ is $\frac{83.2}{100}\times 56.7\times \frac{208.24}{137.33}=71.53\approx 72$ g.