Question

What is the mass of copper produced if 4.87 g of `Al` reacts with copper sulfate in the reaction `2Al + 3CuSO_4 -> Al_2(SO_4)_3 + 3Cu`?

Answer 1

Approx. `17` `g` copper metal result.

Explanation:

`2Al + 3CuSO_4 rarr Al_2(SO_4)_3 + 3Cu`

`"Moles of aluminum " =(4.87*g)/(26.98*g*mol^-1)` `=` `0.181` `mol`

From the equation, we know that `3` moles of copper are reduced per `2` mol aluminum oxidized.

So `3/2xx0.181*mol xx 63.55*g*mol^-1` `=` `?g`

You have given the stoichiometry. It explicitly says that 3 moles of copper salt are reduced per 2 moles of aluminum oxidized. So simply multiply the molar quantity of aluminum by `3/2`.