$n_{SO_2}=20,16/22,4=0,9mol$
$a/PTHH :$
$2Fe+6H_2SO_4\overset{t^o}\to Fe_2(SO_4)_3+6H_2O+3SO_2↑$
$Cu+2H_2SO_4\overset{t^o}\to 2H_2O+SO_2↑+CuSO_4$
$\text{b/Gọi}$ $n_{Fe}=a;n_{Cu}=b$
$\text{Ta có :}$
$m_{hh}=56a+64b=38,6g$
$n_{SO_2}=1,5a+b=0,9mol$
$\text{Ta có hpt :}$
$\left\{\begin{matrix}
56a+64b=38,6& \\
1,5a+b=0,9&
\end{matrix}\right.$
$⇔\left\{\begin{matrix}
a=0,475 & \\
b=0,1875 &
\end{matrix}\right.$
$⇒\%m_{Fe}=\dfrac{0,475.56.100\%}{38,6}=68,91\%$
$⇒\%m_{Cu}=100\%-68,91\%=31,09\%$