CuO+H2SO4--->CuSO4+H2O
b) n CuO=1,6/80=0,02(mol)
m H2SO4=100.20/100=20(g)
n H2SO4=20/98=0,2(mol)
--->H2SO4 dư
dd sau pư là H2SO4 dư và CuSO4
m dd sau pư=1,6+100=101,6(g)
n H2SO4=n CuO=0,02(mol)
m H2SO4=0,02.98=1,96(g)
m H2SO4 dư=20-1,96=18,04(g)
C% H2SO4 dư=18,04/101,6.100%=17,76%
n CuSO4=n CuO=0,02(mol)
m CuSO4=0,02.160=3,2(g)
C% CuSO4=3,2/101,6.100%=3,15%