Hướng dẫn:

${{m}_{c{{t}_{HCl}}}}=\frac{{{m}_{d{{d}_{HCl}}}}.C%}{100}=\frac{500.1,095}{100}=5,475\left( g \right)$ 

$\Rightarrow {{n}_{HCl}}=\frac{{{m}_{HCl}}}{{{M}_{HCl}}}=\frac{5,475}{36,5}=0,15\left( mol \right)$ 

$Fe+2HCl\to FeC{{l}_{2}}+{{H}_{2}}$ 

           0,15$\to $               0,075 (mol)

Theo phương trình ta có:

${{n}_{{{H}_{2}}}}=\frac{{{n}_{HCl}}}{2}=\frac{0,15}{2}=0,075\left( mol \right)$ 

${{V}_{{{H}_{2}}}}=22,4.{{n}_{{{H}_{2}}}}=22,4.0,075=1,68\left( l \right)$ 

Chọn đáp án A