Đáp án:

$H=93\%$

Giải thích các bước giải:

$Fe+S\xrightarrow{t^o}FeS$

$Fe+2HCl\to FeCl_2+H_2↑$

$FeS+2HCl\to FeCl_2+H_2S↑$

$\overline{M}_Z=13,4.2=26,8(g/mol)$

$\to \frac{n_{H_2}}{n_{H_2S}}=\frac{34-26,8}{26,8-2}=\frac{9}{31}$

Mà $n_{H_2}+n_{H_2S}=\frac{2,688}{22,4}=0,12$

$\to n_{H_2}=0,027(mol);n_{H_2S}=0,093(mol)$

$n_S=\frac{3,2}{32}=0,1(mol)$

Theo PT: $n_{Fe(pứ)}=n_{S(pứ)}=n_{FeS}=n_{H_2S}=0,093(mol)$

$\to H=\frac{0,093}{0,1}.100\%=93\%$

Đáp án:

$H=93\%$ 

Giải thích các bước giải:

$n_{S}=\dfrac{3,2}{32}=0,1(mol)$

$Fe+S\xrightarrow{t^o}FeS$

$FeS+2HCl\to FeCl_2+H_2S$

$Fe+2HCl\to FeCl_2+H_2$

$\overline{M}_Z=13,4×2=26,8(g/mol)$

$n_{Z}=\dfrac{2,688}{22,4}=0,12(mol)$

$\dfrac{n_{H_2S}}{n_{H_2}}=\dfrac{26,8-2}{34-26,8}=\dfrac{31}{9}$

$n_{H_2S}=\dfrac{31}{40}×0,12=0,093(mol)$

$n_{H_2}=0,12-0,093=0,027(mol)$

Bảo toàn nguyên tố $S:$

$n_{H_2S}=n_{S\ \rm (phản\ ứng)}=0,093(mol)$

$H=\dfrac{0,093}{0,1}×100\%=93\%$