Hướng dẫn:

${{n}_{{{H}_{2}}}}=\frac{{{V}_{{{H}_{2}}}}}{22,4}=\frac{3,36}{22,4}=0,15\left( mol \right)$ 

${{n}_{{{H}_{2}}S}}=\frac{{{V}_{{{H}_{2}}S}}}{22,4}=\frac{2,24}{22,4}=0,1\left( mol \right)$

$Fe+2HCl\to FeC{{l}_{2}}+{{H}_{2}}\uparrow $         (1)

          0,3   $\leftarrow $           0,15        (mol)

$FeS+2HCl\to FeC{{l}_{2}}+{{H}_{2}}S\uparrow $       (2)

            0,2  $\leftarrow $         0,1

Theo phương trình(1) và (2) ta có:

${{n}_{HCl}}=2.{{n}_{{{H}_{2}}}}+2.{{n}_{{{H}_{2}}S}}=0,15.2+2.0,1=0,5\left( mol \right)$ 

${{V}_{HCl}}=\frac{{{n}_{HCl}}}{{{C}_{{{M}_{HCl}}}}}=\frac{0,5}{2}=0,25\left( l \right)=250ml$  

Chọn đáp án B