Gọi a, b là mol Cr, Mg
=> $52a+24b= 2$ (1)
$n_{NO_2}= 0,14 mol$
$Cr+ 6HNO_3 \to Cr(NO_3)_3+ 3NO_2+ 3H_2O$
$Mg+ 4HNO_3 \to Mg(NO_3)_2+ 2NO_2+ 2H_2O$
=> $3a+2b=0,14$ (2)
(1)(2) => $a=0,02; b=0,04$
$n_{kim loại}= n_{muối}$
=> $m= 0,02.238+148.0,04= 10,68g$
$\%Cr= \frac{52.0,02.100}{2}= 52\%$