a) \(n_{CuSO_4}=\dfrac{32.10\%}{160}=0,02\left(mol\right)\)
PTHH: Zn + CuSO4 ---> ZnSO4 + Cu (Phản ứng thế)
0,02<--0,02----->0,02------>0,02
b) mZn (pư) = 0,02.65 = 1,3 (g)
c) mdd = 32 + 0,02.65 - 0,02.64 = 32,02 (g)
=> \(C\%_{ZnSO_4}=\dfrac{0,02.161}{32,02}.100\%=10,056\%\)