nCuO=1,6/80=0,02(mol)
PTHH: CuO + H2SO4 -> CuSO4 + H2O
nCuSO4=nCuO=0,02(mol)->mCuO=0,02.160= 3,2(g)
mddCuSO4=mCuO + mddH2SO4=1,6+ 100=101,6(g)
=>C%ddCuSO4=(3,2/101,6).100=3,15%
nCuO = 0,02 (mol)
Bảo toàn Cu => nCuO = nCuSO4 = 0,02 (mol)
=> C% dd = 0,02.160/1,6+100 . 100% = 3,15^