CuO+H2SO4--->CuSO4+H2O
m H2SO4=100.20/100=20(g)
n H2SO4=20/98\(\approx0,2\left(mol\right)\)
n CuO=1,6/80=0,02(mol)
-->CuO hết
n H2SO4 dư=0,2-0,02=0,18(g)
m H2SO4=0,18.98=17,64(g)
m dd sau pư=100+1,6=101,6(g)
dd sau pưu gồm H2SO4 du2 và CuSO4
C% H2SO4=17,64/101,6.100%=17,36%
n CuSO4=n CuO=0,02(mol)
m CuSO4=0,02.160=3,2(g)
C% CuSO4=3,2/101,6.100%=3,15%