$m_{giảm}= m_{X pứ}- m_{H_2}= 32.13,125\%= 4,2g$
$n_{H_2}= \frac{3,92}{22,4}= 0,175 mol$
$\Rightarrow m_{X pứ}= m_{H_2}+ 4,2= 0,175.2+4,2= 4,55g$
$2X+2nHCl \to 2XCl_n+ nH_2$
$\Rightarrow n_X=\frac{0,35}{n} mol$
$\Rightarrow M_X=\frac{4,55n}{0,35}= 13n$
(không có n thoả mãn M).