# Based on the following equation: 2 Al + 3 CuCl_2 -> 2AlCl_3 + 3 Cu. lf 10.53 grams of CuCl_2 were reacted, how many grams of AlCl_3 will be produced?

Feb 18, 2016

The law of equivalent proportion may help solving the problem.

#### Explanation:

According to this law a reactant will produce same no of equivalent of product/s as its reacting mass has.
here the equivalent mass of $C u C {l}_{2}$ = Formula wt /total valency of metal(Cu) = $\frac{63.55 + 2 \cdot 35.55}{1 \cdot 2}$ =67.33
Again
the equivalent mass of $A l C {l}_{3}$ = Formula wt /total valency of metal(Al) = $\frac{27 + 3 \cdot 35.55}{1 \cdot 3}$ =44.55
Now no. of gram equivalent in 10.53g of $C u C {l}_{2}$ reacted = $\frac{10.53}{67.33}$=0.156
hence the same no.of gram equivalent of $A l C {l}_{3}$ will be produced
So the mass of $A l C {l}_{3}$ produced =0.156x equivalent mass of $A l C {l}_{3}$ 0.156x44.55 = 6.95 g

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