calculate the percent yield if 1.22 g of copper is produced from 5.60 g of copper(II) sulfate and excess zinc. answer to three sig figs Zn(s) + CuSO4(aq) --> ZnSO4(aq) + Cu(s)
calculate the percent yield if 1.22 g of copper is produced from 5.60 g of copper(II) sulfate and excess zinc. answer to three sig figs
Zn(s) + CuSO4(aq) --> ZnSO4(aq) + Cu(s)
Given reaction is
Zn(s) + CuSO4(aq) --> ZnSO4(aq) + Cu(s)
as per equation 1mol 1 mol 1 mol 1 mol
= 159.5 g 63.5 g
as per problem 5.60 g ?
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