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 $100$.