Đặt $n_S=x(mol);n_{Mg}=y(mol)(x;y>0)$Ta có $m_X=13,6(g)\Leftrightarrow 32x+24y=13,6(1)$
PT:$ S+Mg\rightarrow MgS$
x---->x ---->x (mol)
MgS+2HCl----->MgCl2+H2S
x----> x (mol)
Mg+2HCl----->MgCl2+H2
(y-x)---> (y-x) (mol)
$m_Z=\frac{6,72}{22,4}=0,3(mol)\Leftrightarrow y=0,3(2)$
Thay (2) vào (1)$\Rightarrow \left\{ \begin{array}{l} x=0,2\\ y=0,3 \end{array} \right.\Rightarrow \frac{n_{H_2}}{n_{H_{2}S}}=\frac{1}{2}\Leftrightarrow \frac{34-2a}{2a-2}=\frac{1}{2}\Leftrightarrow a=\frac{35}{3}$
Vậy ........