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

Md Ariful Islam

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Md Ariful Islam

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

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

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 $3/2$.

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