# 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?

Apr 12, 2016

Approx. $17$ $g$ copper metal result.

#### Explanation:

$2 A l + 3 C u S {O}_{4} \rightarrow A {l}_{2} {\left(S {O}_{4}\right)}_{3} + 3 C u$

$\text{Moles of aluminum } = \frac{4.87 \cdot g}{26.98 \cdot g \cdot m o {l}^{-} 1}$ $=$ $0.181$ $m o l$

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

So $\frac{3}{2} \times 0.181 \cdot m o l \times 63.55 \cdot g \cdot m o {l}^{-} 1$ $=$ ?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 $\frac{3}{2}$.