${{n}_{CuO}}=\frac{{{m}_{CuO}}}{{{M}_{CuO}}}=\frac{12}{80}=0,15\left( mol \right)$

${{n}_{MgO}}=\frac{{{m}_{MgO}}}{{{M}_{MgO}}}=\frac{8}{40}=0,2\left( mol \right)$

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

0,15$\leftarrow $0,15                      (mol)

$CO+MgO\xrightarrow{{{t}^{0}}}Mg+C{{O}_{2}}$

0,2$\leftarrow $0,2                             (mol)

Theo 2 phương trình ta có:

$\sum{{{n}_{CO}}={{n}_{CuO}}+{{n}_{MgO}}=0,15+0,2=0,35\left( mol \right)}$

$\Rightarrow {{V}_{CO}}=22,4.\sum{{{n}_{CO}}}=22,4.0,35=7,84\left( ml \right)$

$\Rightarrow $Chọn đáp án A