Bạn tham khảo:
$a/ CuO+H_2O_4 \to CuSO_4+H_2O$
$b/$
$n_{H_2SO_4}=\frac{100.19,6\%}{98}=0,2(mol)$
$n_{CuO}=\frac{8}{80}=0,1(mol)$
$CuO$ hết, $H_2SO_4$ dư
$m_{H_2SO_4(dư)}=0,1.98=9.8(g)$
$c/$
$C\%_{CuSO_4}=\frac{0,1.160}{8+100}.100\%=14,8\%$
$C\%_{H_2SO_4}=\frac{98}{8+100}.100\%=9\%$