$Mg+2HCl\to MgC{{l}_{2}}+{{H}_{2}}$

    0,1           $\to $  0,1              (mol)

   $MgO+2HCl\to MgC{{l}_{2}}+{{H}_{2}}O$

    0,1               $\to $ 0,1               (mol)

$\begin{align} & {{n}_{Mg}}=\frac{{{m}_{Mg}}}{{{M}_{Mg}}}=\frac{2,4}{24}=0,1\left( mol \right) \\& {{n}_{MgO}}=\frac{{{m}_{MgO}}}{{{M}_{MgO}}}=\frac{6}{40}=0,15\left( mol \right) \\\end{align}$

Theo phương trình ta có: $\begin{align}  & {{n}_{MgC{{l}_{2}}}}={{n}_{Mg}}+{{n}_{MgO}}=0,1+0,15=0,25\left( mol \right) \\& {{m}_{MgC{{l}_{2}}}}={{n}_{MgC{{l}_{2}}}}.{{M}_{MgC{{l}_{2}}}}=0,25.95=23,75\left( mol \right) \\\end{align}$

$\Rightarrow $Chọn đáp án B