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#?

1 Answer
anor277
Apr 12, 2016

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#.

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