Đáp án:
$Mg+4HNO_3→Mg(NO_3)_2+2NO_2+2H_2O$
a 2a
$Fe+6HNO_3→Fe(NO_3)_3+3NO_2+3H_2O$
b 3b
$n_{NO_2}=$$\dfrac{11,2}{22,4}=0,5(mol)$
Ta có hpt:
$\begin{cases}24a+56b=8\\2a+3b=0,5\\\end{cases}$ $⇔a=0,1; b=0,1$
$⇒n_{Mg}=0,1(mol)⇒m_{Mg}=0,1.24=2,4(g)⇒$%$_{Mg}=$$\dfrac{2,4}{8}.100=30$%
$⇒$%$_{Fe}=70$%
BẠN THAM KHẢO NHA!!!