$m_{dd}=45.1,12=50,4g$
$\Rightarrow n_{CuSO_4}=\dfrac{50,4.25\%}{160}=0,07875(mol)$
$\Delta m=7,24-7=0,24g=m_{Cu}-m_{Fe\text{pứ}}$
$Fe+CuSO_4\to FeSO_4+Cu$
Đặt $n_{Fe\text{pứ}}=n_{CuSO_4\text{pứ}}=n_{FeSO_4}=n_{Cu}=x$
$\Rightarrow 64x-56x=0,24$
$\Leftrightarrow x=0,03$
$\Rightarrow n_{CuSO_4\text{dư}}=0,07875-x=0,04875(mol)$
$m_{dd\text{spứ}}=56x+50,4-64x=50,16g$
$m_{CuSO_4}=0,04875.160=7,8g\Rightarrow C\%_{CuSO_4}=15,55\%$
$m_{FeSO_4}=152.0,03=4,56g\Rightarrow C\%_{FeSO_4}=9,1\%$