Question

If 1.10 g of Zn is allowed to react with 2.70 g of CuSO4, according to the equation below, how many grams of Zn will remain after the reaction is complete?

CuSO4(aq)+Zn(s)→ZnSO4(aq)+Cu(s)

Answer 1

Initial amounts of Zn and CuSO₄ in moles should be determined first:

$n(Zn)=\frac{1.10g}{65.38g/mol}=0.0168mol$

$n(CuSO_4)=\frac{2.70g}{159.61g/mol}=0.0169mol$

According to the equation, Zn and CuSO₄ react in mole ratio of 1 : 1. So there will be no Zn remaining after the reaction as it is a limiting reactant and will be used up completely.

Answer: no Zn will remain at all