Đáp án:

$\rm a)\\n_{Fe}=\dfrac{16,8}{56}=0,3\ (mol)\\n_S=\dfrac{6,8}{32}=0,2125\ (mol)$

Phương trình phản ứng:

$\rm Fe+S\xrightarrow{t^°}FeS$

Ta có: $\rm n_{Fe}>n_S\to Fe$ dư sau phản ứng.

$\rm Fe+2HCl\to FeCl_2+H_2\\FeS+2HCl\to FeCl_2+H_2S$

$\rm b)$

Theo phương trình:

$\rm n_{FeS}=n_{H_2S}=n_S=0,2125\ (mol)$

$\rm n_{Fe\ (du)}=n_{H_2}=0,3-0,2125=0,0875\ (mol)$

$\rm n_{HCl}=2(n_{FeS}+n_{Fe\ (du)})=0,6\ (mol)$

$\rm\to V_{HCl}=\dfrac{0,6}{1}=0,6\ (M)$

$\rm c)$

$\rm V_B=22,4×(0,2125+0,0875)=6,72\ (L)$

$\rm d_{B/CH_4}=\dfrac{0,2125×34+0,0875×2}{12+4}=0,4625$