a)
$CuO + H_2SO_4 \to CuSO_4 + H_2O$
$CuO + 2HCl \to CuCl_2 + H_2O$
b) $n_{H_2SO_4} = 0,15V(mol) ; n_{HCl} = 1,7V(mol)$
Theo PTHH :
$n_{CuO} =\dfrac{1}{2}n_{HCl} + n_{H_2SO_4} = V = \dfrac{15}{80} = 0,1875(mol)$
$\Rightarrow V = 0,1875(lít)$
$n_{CuSO_4} = n_{H_2SO_4} = 0,028125(mol)$
$n_{CuCl_2} = \dfrac{1}{2}n_{HCl} = 0,159375(mol)$
$\Rightarrow m_{muối} = 0,028125.160 + 0,159375.135 = 26,015625(gam)$