Hướng dẫn:

${{n}_{Cu}}=\frac{{{m}_{Cu}}}{{{M}_{Cu}}}=\frac{3,2}{64}=0,05\left( mol \right)$

$Cu+S\xrightarrow{{{t}^{0}}}CuS$  (1)

0,05       $\to $ 0,05                    (mol)

$CuS+2HCl\to CuC{{l}_{2}}+{{H}_{2}}S\uparrow $         (2)

0,05           $\to $           0,05                                  (mol)

Theo phương trình (1) có :${{n}_{Cu}}={{n}_{CuS}}=0,05\left( mol \right)$

Theo phương trình (2) có: ${{n}_{CuS}}={{n}_{{{H}_{2}}S}}=0,05\left( mol \right)$

$\Rightarrow {{V}_{{{H}_{2}}S}}=22,4.{{n}_{{{H}_{2}}S}}=22,4.0,05=1,12\left( l \right)$

Chọn đáp án B