-Gọi $n_{Fe(OH)_3}=x(mol)$
$n_{Cu(OH)_2}=y(mol)$
$⇒107x+98y=3,61(g)(1)$
-Bảo toàn $Fe$ ⇒$n_{Fe_2O_3}=\frac{1}{2}.n_{Fe(OH)_3}=\frac{1}{2}.x(mol)$
-Bảo toàn $Cu$ ⇒$n_{CuO}=n_{Cu(OH)_2}=y(mol)$
$⇒\frac{1}{2}.160x+80y=2,8(g)$
$⇔80x+80y=2,8(g)(2)$
-Từ (1) và (2),ta có hệ pt:
$\left \{ {{107x+98y=3,61} \atop {80x+80y=2,8}} \right.$ $\left \{ {{x=0,02} \atop {y=0,015}} \right.$
$⇒m_{Fe(OH)_3}=0,02.107=2,14(g)$
$⇒m_{Cu(OH)_2}=0,015.98=1,47(g)$