If $3.7~\mathrm{mol}$ of $\ce{Cu}$ reacts with an excess of $\ce{Cl2}$ and the exchange is $45\,\%$, how many moles of $\ce{CuCl2}$ can at maximum be created?

So I start with write the reaction formula $$\ce{Cu + Cl2 -> CuCl2}$$ and then balance it. $$\ce{2Cu + 2Cl2 -> 2CuCl2}$$ I know that I have $3.7\ \mathrm{mol}$ of $\ce{Cu}$, which is $235.1202\ \mathrm{g}$ of $\ce{Cu}$. I assume that $\ce{Cu}$ is limiting reagent and $\ce{Cl2}$ is an excess reagent.

This is where I get stuck, I don’t know how I’m supposed to proceed with the only information that the exchange is $45\,\%$.

  • $\begingroup$ Your "balancing" consisted of multiplying the whole equation by 2. Your original equation was balanced $\endgroup$ – orthocresol Jun 12 '15 at 20:18

We have this equation of reaction:

$$\ce{Cu + Cl2 -> CuCl2}$$

Which means that $1~\mathrm{mol}$ of $\ce{Cu}$ reacts with $1~\mathrm{mol}$ of $\ce{Cl2}$ to form $1~\mathrm{mol}$ of $\ce{CuCl2}$.

Therefore, we would expect $3.7~\mathrm{mol}$ of $\ce{Cu}$ to produce $3.7~\mathrm{mol}$ of $\ce{CuCl2}$ if there is excess $\ce{Cl2}$, which is stated in the question.

However, the exchange is only $45\%$, which means that only $45\%$ of $\ce{Cu}$ is converted, making we can only create $3.7\times45\%=1.665~\mathrm{mol}$ of $\ce{CuCl2}$.

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