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
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