$PTHH : Fe +S \xrightarrow{to} FeS$ (1)

$FeS +2HCl \xrightarrow{to} FeCl_2 +H_2S$ (2)

$n_{H_2S}=\dfrac{4,48}{22,4}=0,2(mol)$ 

Theo pt(2)  $n_{FeS}=n_{H_2S}=0,2(mol)$ 

Theo pt (2) $n_{Fe}=n_{S}=n_{FeS}=0,2(mol)$ 

$m_{Fe}=0,2 ×56=11,2(g)$ 

%$m_{Fe}=\dfrac{11,2}{14,4}×100$%$≈77,78$%

%$m_{S}=100-77,78 =22,22$%

$9.$

$a.$

$Fe+S \xrightarrow{t^o}FeS$ $(1)$

$FeS+2HCl \to FeCl_2+H_2S$ $(2)$

$Fe_{dư}+2HCl \to FeCl_2+H_2$ $(3)$

$b.$

$n_{Fe}=a(mol);n_{S}=b(mol)$

`->`$56a+32b=14.4$ $(*)$

$n_{H_2S+H_2}=\dfrac{4.48}{22.4}=0.2(mol)$

Từ phương trình phản ứng $(2),(3)$ có:

$n_{Fe}=n_{H_2S}+n_{H_2}$

`->`$a=0.2(mol)$

Từ $(*)$ suy ra $b=0.1$

`->`$\%m_{Fe}=\dfrac{0.2 \times 56}{14.4} \times 100\%=77.78\%$

`->`$\%m_{S}=\dfrac{0.1 \times 32}{14.4} \times 100\%=22.22\%$