Đặt: $n_{Fe}=x\ (mol)$; $n_{Al}=y\ (mol)$
$\to n_{Fe(NO_3)_3}=x\ (mol)$; $n_{Al(NO_3)_3}=y\ (mol)$
$\to 242x+213y=69,7\ \ (1)$
$Fe+4HNO_3\to Fe(NO_3)_3+NO+2H_2O$
$Al+4HNO_3\to Al(NO_3)_3+NO+2H_2O$
$n_{NO}=\dfrac{6,72}{22,4}=0,3\ (mol)$
$\to x+y=0,3\ \ (2)$
Từ $(1)$ và $(2)$ suy ra: $x=0,2\ (mol)$ và $y=0,1\ (mol)$
$\to \left\{\begin{matrix}m_{Fe}=0,2.56=11,2\ (g)\\m_{Al}=0,1.27=2,7\ (g)\end{matrix}\right.$