Stoichiometry of the redox reaction with crystallohydrate [closed]








**Closed.** This question is [off-topic](/help/closed-questions). It is not currently accepting answers.
 
 






---








 **Homework questions** must demonstrate some effort to understand the underlying concepts. For help asking a good homework question, see: [How do I ask homework questions on Chemistry Stack Exchange?](https://chemistry.meta.stackexchange.com/questions/141/how-do-i-ask-homework-questions-on-chemistry-stack-exchange)




Closed 4 years ago.








[Improve this question](/posts/111357/edit) 




> 
> Consider the following balanced equation:
> 
> 
> $\ce{2 Al(s) + 3 CuCl2 . 2 H2O (aq) → 3 Cu (s) + 2 AlCl3 (aq) + 6 H2O (l)}$
> 
> 
> If we began the experiment with 0.99 g of Al, according to the stoichiometry of the reaction how much $\ce{CuCl2.2 H2O}$ should be used to complete the reaction without either reactant being in excess?
> 
> 
> Answer the following using the correct number of significant figures.
> 
> 
> 


I cannot for the life of me solve this problem. Should I be finding the mole of $\ce{Al}$ and $\ce{CuCl2.2H2O}$ then find the difference? Any help would be appreciated.




Mass of $\ce{CuCl2.2H2O}$ needed can be found from its amount $n(\ce{CuCl2.2H2O})$ and molar mass $M(\ce{CuCl2.2H2O})$:


$$m(\ce{CuCl2.2H2O}) = n(\ce{CuCl2.2H2O})\cdot M(\ce{CuCl2.2H2O}) \label{eqn:1}\tag{1}$$


The unknown amount of copper(II) chloride dihydrate $n(\ce{CuCl2.2H2O})$ can be found from the reaction stoichiometry:


$$n(\ce{CuCl2.2H2O}) = \frac{3\cdot n(\ce{Al})}{2} = \frac{3\cdot m(\ce{Al})}{2\cdot M(\ce{Al})} \label{eqn:2}\tag{2}$$


Finally, plugging $\eqref{eqn:2}$ into $\eqref{eqn:1}$:


$$
\begin{align}
m(\ce{CuCl2.2H2O}) &= \frac{3\cdot M(\ce{CuCl2.2H2O})\cdot m(\ce{Al})}{2\cdot M(\ce{Al})} \\
 &= \frac{3\cdot\pu{170.5 g mol-1}\cdot\pu{0.99 g}}{2\cdot\pu{26.98 g mol-1}} \\
 &\approx \pu{9.40 g}
\end{align}
$$


Hence, you need $\pu{9.40 g}$ of $\ce{CuCl2.2H2O}$.


 2