a, 

B gồm $H_2$ (x mol), $H_2S$ (y mol)

$n_B=\dfrac{6,72}{22,4}=0,3(mol)$

$\Rightarrow x+y=0,3$                 (1)

$\overline{M}_B=11,666.2=23,332$

$\Rightarrow 2x+34y=23,332.0,3=7$   (2)

(1)(2)$\Rightarrow x=0,1; y=0,2$

$\%V_{H_2}=\dfrac{0,1.100}{0,3}=33,33\%\Rightarrow \%V_{H_2S}=66,67\%$

b,

$M+S\buildrel{{t^o}}\over\to MS$

$MS+2HCl\to MCl_2+H_2S$

$M+2HCl\to MCl_2+H_2$

$n_{M\text{dư}}=n_{H_2}=0,1(mol)$

$n_S=n_{M\text{pứ}}=n_{H_2S}=0,2(mol)$

$\Rightarrow 0,2.32+M(0,2+0,1)=25,9$

$\Leftrightarrow M=65(Zn)$

a.

-Gọi $n_{H_2}=x (mol)$

       $n_{H_2S}=y(mol)$

$⇒n_{hh...B}=x+y=\frac{6,72}{22,4}=0,3(mol)(1)$ 

-Vì $B$ có tỉ khối với $H_2$ là 11,666

$⇒\frac{m_{H_2}.x+m_{H_2S}.y}{x+y}=\frac{2x+34y}{x+y}=11,666.2=23,332$

$⇔-21,332x+10,668y=0(2)$

-Từ (1) và (2),ta có hệ pt:

$\left \{ {{x+y=0,3} \atop {-21,332x+10,668y=0}} \right.$ $\left \{ {{x≈0,1} \atop {y≈0,2}} \right.$

-Vì %$V=$%$n$

%$V_{H_2}=$%$n_{H_2}=\frac{0,1}{0,3}.100$ %$≈33,33$%

%$V_{H_2S}=100$%$-$%$V_{H_2}=100$%$-33,33$%$=66,67$%

b.

Bảo toàn $S$ ⇒$n_S=n_{H_2S}=0,2(mol)$

Mà $n_M=n_S+n_{H_2S}=0,1+0,2=0,3(mol)$

⇒Ta có: $m_S+m_M=25,9(g)$

    $⇔0,2.32+0,3.M_M=25,9$

    $⇔M_M=65(g/mol)$

⇒$M$ là nguyên tố Kẽm ($Zn$)