Hướng dẫn:

${{n}_{{{H}_{2}}}}=\frac{{{V}_{{{H}_{2}}}}}{22,4}=\frac{5,6}{22,4}=0,25\left( mol \right)$

  ${{H}_{2}}+CuO\xrightarrow{{{t}^{0}}}Cu+{{H}_{2}}O$

0,25$\to $0,25                              (mol)

Theo phương trình ta có: ${{n}_{{{H}_{2}}}}={{n}_{CuO}}=0,25\left( mol \right)$

$\Rightarrow {{m}_{CuO}}={{n}_{CuO}}.{{M}_{CuO}}=0,25.80=20\left( g \right)$

$\Rightarrow $ Chọn đáp án A