Hướng dẫn:

$\begin{align}  & {{n}_{C{{O}_{2}},S{{O}_{2}}}}=\frac{{{V}_{hh}}}{22,4}=\frac{5,6}{22,4}=0,25 \\ & C{{O}_{2}}+Ba{{(OH)}_{2}}\to BaC{{O}_{3}}+{{H}_{2}}O(i) \\ & a\to a \\ & S{{O}_{2}}+Ba{{(OH)}_{2}}\to BaS{{O}_{3}}+{{H}_{2}}O(ii) \\ & b\to b \\\end{align}$

Theo phương trình (i) và (ii) ta có: ${{n}_{Ba{{(OH)}_{2}}}}=a+b={{n}_{C{{O}_{2}}}}+{{n}_{S{{O}_{2}}}}=0,25$ (mol)

$x={{C}_{{{M}_{Ba{{(OH)}_{2}}}}}}=\frac{{{n}_{Ba{{(OH)}_{2}}}}}{{{V}_{Ba{{(OH)}_{2}}}}}=\frac{0,25}{0,4}=0,625$ (M)

                $\Rightarrow $ Chọn đáp án D.