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.


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    $\begingroup$ HINT - You need convert to convert grams into something that is proportional to the stoichiometric values. $\endgroup$ – MaxW Mar 22 at 4:18
  • $\begingroup$ Of all arithmetic operations, subtraction is the only one you won't be needing here. $\endgroup$ – Ivan Neretin Mar 22 at 4:48

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 $\ce{CuCl2.2H2O}$.

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    $\begingroup$ You made a bad rounding error. With a modern calculator you really should carry at least two extra digits in intermediate calculations and only round the final result to the appropriate number of significant figures. $\endgroup$ – MaxW Mar 22 at 8:55
  • $\begingroup$ @MaxW is right. What I would suggest is not to do any intermediate calculations at all, unless you are explicitly asked so. Solve the problem algebraically first, plug in the numbers at the very last step. Also, note that "number of moles" is a colloquial expression for the amount of substance and should be avoided. It's like replacing mass and voltage with "number of kilograms" number "number of volts", respectively. $\endgroup$ – andselisk Mar 22 at 9:56
  • $\begingroup$ What "now"? There is still "No. of moles". Additionally to my last comment, note that algebraically the amount is usually denoted as $n(\ce{A_xB_y})$ — please don't write an essay in math formulas. If you want, I may try to rewrite your answer to better meet the standard notations. $\endgroup$ – andselisk Mar 22 at 10:02
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    $\begingroup$ I've rewritten your answer from ground up preserving your logic (which is correct), addressing the issues regarding the notations, units and significant figures. Simply put, this is how I'd answer this. If you are unhappy with this, feel free to revert my edit; otherwise have a closer look at the formatting and used notations so that you could improve your future posts:) $\endgroup$ – andselisk Mar 22 at 10:26
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    $\begingroup$ Thanks a lot @andselisk $\endgroup$ – Fitz Watson Mar 22 at 10:44

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