Hướng dẫn:

${{n}_{HCl}}={{V}_{HCl}}.{{C}_{{{M}_{HCl}}}}=0,2.0,5=0,1\left( mol \right)$

$2Cu+{{O}_{2}}\xrightarrow{{{t}^{0}}}2CuO$       (1)

0,05              0,05                    (mol)

$CuO+2HCl\to CuC{{l}_{2}}+{{H}_{2}}O$    (2)

0,05  0,1                                   (mol)

Theo phương trình (2) có: ${{n}_{CuO}}=\frac{{{n}_{HCl}}}{2}=\frac{0,1}{2}=0,05\left( mol \right)$

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

$\Rightarrow {{m}_{Cu}}={{n}_{Cu}}.{{M}_{Cu}}=0,05.64=3,2\left( g \right)$

$\Rightarrow $Chọn đáp án B