Question

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?

Answer 1

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 `CuCl_2` = Formula wt /total valency of metal(Cu) = `(63.55+2*35.55)/(1*2)` =67.33
Again
the equivalent mass of `AlCl_3` = Formula wt /total valency of metal(Al) = `(27+3*35.55)/(1*3)` =44.55
Now no. of gram equivalent in 10.53g of `CuCl_2` reacted = `10.53/67.33`=0.156
hence the same no.of gram equivalent of `AlCl_3` will be produced
So the mass of `AlCl_3` produced =0.156x equivalent mass of `AlCl_3` 0.156x44.55 = 6.95 g

Pl, give feed back