$n_{SO_2}=5,6/22,4=0,25mol$
$2Fe+6H_2SO_4\overset{t^o}\to Fe_2(SO_4)_3+6H_2O+3SO_2↑$
$Cu+2H_2SO_4\overset{t^o}\to 2H_2O+SO_2↑+CuSO_4$
Gọi $n_{Fe}=a;n_{Cu}=b$
Ta có :
$m_{hh}=56a+64b=12$
$n_{SO_2}=1,5a+b=0,25$
Ta có hpt :
$\left\{\begin{matrix}
56a+64b=12 & \\
1,5a+b=0,25 &
\end{matrix}\right.$
$⇔\left\{\begin{matrix}
a=0,1 & \\
b=0,1 &
\end{matrix}\right.$
$\%m_{Fe}=\dfrac{0,1.56.100\%}{12}=46,667\%$
$⇒\%m_{Cu}=100\%-46,667\%=53,333\%$