a) $CuSO_4 + 2NaOH \to Cu(OH)_2 + Na_2SO_4$
$n_{NaOH} = \dfrac{200.6\%}{40} =0,3(mol)$
Theo PTHH, $n_{CuSO_4} = \dfrac{1}{2}n_{NaOH} = 0,15(mol)$
$\Rightarrow m_{dd\ CuSO_4} = \dfrac{0,15.160}{16\%} = 150(gam)$
b) $n_{Cu(OH)_2} = n_{CuSO_4} =0,15(mol)$
$m_{dd\ sau\ pư} = m_{dd\ NaOH} + m_{dd\ CuSO_4} - m_{Cu(OH)_2}$
$= 200 + 150 - 0,15.98 = 335,3(gam)$
$n_{Na_2SO_4} = 0,15(mol)$
$\Rightarrow C\%_{Na_2SO_4} = \dfrac{0,15.142}{335,3}.100\% = 6,35\%$