a)\ CuO+H_2SO_4->CuSO_4+H_2O

b)

Theo PT: n_{CuO}=n_{H_2SO_4}=n_{CuSO_4}={1,6}/{80}=0,02(mol)

->m_{H_2SO_4\ du}=100.20\%-0,02.98=18,04(g)

m_{dd\ spu}=1,6+100=101,6(g)

->C\%_{H_2SO_4\ du}={18,04}/{101,6}.100\%\approx 17,76\%

       C\%_{CuSO_4}={0,02.160}/{101,6}.100\%\approx 3,15\%