Hướng dẫn giải

- Chọn D.

- Ta có: ${{n}_{C{{O}_{2}}}}=\frac{1,68}{22,4}=0,075$ (mol)

- Gọi x, y lần lượt là số mol của${{A}_{2}}C{{O}_{3}}$và $BC{{O}_{3}}$.

- PTHH:          ${{A}_{2}}C{{O}_{3}}$+  2HCl$\xrightarrow{{}}$  2ACl  +  $\text{C}{{\text{O}}_{2}}\uparrow $ +  ${{H}_{2}}O$

   (mol)         x    $\to $    2x           2x            x              x

$BC{{O}_{3}}$+  2HCl$\xrightarrow{{}}$$BC{{l}_{2}}$  +  $\text{C}{{\text{O}}_{2}}\uparrow $ +  ${{H}_{2}}O$

           (mol)        y  $\to $    2y                    y             y               y   

- Theo PTHH: $\left\{\begin{align}&{{n}_{{{H}_{2}}O}}=x+y={{n}_{C{{O}_{2}}}}=0,075(mol) \\  & {{n}_{HCl}}=2x+2y=2{{n}_{C{{O}_{2}}}}=2.0,075=0,15(mol) \\ \end{align} \right.$

$\to \left\{ \begin{align}& {{\text{m}}_{\text{C}{{\text{O}}_{\text{2}}}}}\text{ = 0,075}\text{.44 = 3,3 (gam)} \\&{{\text{m}}_{{{\text{H}}_{\text{2}}}\text{O}}}\text{ = 0,075}\text{.18 = 1,35 (gam)} \\  & {{\text{m}}_{\text{HCl}}}\text{=0,15}\text{.36,5 = 5,475(gam)} \\ \end{align} \right.$

- Áp dụng định luật bảo toàn khối lượng: ${{m}_{X}}+{{m}_{HCl}}={{m}_{Y}}+{{m}_{C{{O}_{2}}}}+{{m}_{{{H}_{2}}O}}$

$\to $${{m}_{Y}}={{m}_{X}}+{{m}_{HCl}}-{{m}_{C{{O}_{2}}}}-{{m}_{{{H}_{2}}O}}$= 5 + 5,475 - 3,3 – 1,35 = 5,825 (gam).