nCuO = 6.4/80=0.08 mol
mHCl = 7.3 g
nHCl = 7.3/36.5=0.2 mol
CuO + 2HCl --> CuCl2 + H2O
Bđ: 0.08___0.2
Pư : 0.08__0.16_____0.08
Kt : 0_____0.04_____0.08
mHCl dư = 0.04*36.5 = 1.46 g
mCuCl2 = 0.08*135=10.8 g
mdd sau phản ứng = 6.4 + 36.5 = 42.9 g
C%HCl dư = 1.46/42.9*100% = 3.4%
C%CuCl2 = 10.8/42.9*100% = 25.17%