Hướng dẫn:

${{n}_{CO}}=\frac{{{V}_{CO}}}{22,4}=\frac{8,96}{22,4}=0,4\left( mol \right)$

${{n}_{CuO}}=\frac{{{m}_{CuO}}}{{{M}_{CuO}}}=\frac{28}{80}=0,35\left( mol \right)$

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

Ban đầu       0,4       0,35                                     (mol)

Phản ứng     0,35$\leftarrow $0,35                       (mol)

Theo phương trính ta có: ${{n}_{C{{O}_{pu}}}}={{n}_{CuO}}=0,35\left( mol \right)$

$\Rightarrow %H=\frac{{{n}_{C{{O}_{pu}}}}}{{{n}_{CO}}}.100=\frac{0,35}{0,4}.100=87,5\left( % \right)$

$\Rightarrow $Chọn đáp án B