Giả sử hai kim loại là M (hóa trị II) và B (hóa trị III)
a)
$MC{{O}_{3}}+2HCl\to MC{{l}_{2}}+C{{O}_{2}}+{{H}_{2}}O$
${{B}_{2}}{{(C{{O}_{3}})}_{3}}+6HCl\to 2BC{{l}_{3}}+3C{{O}_{2}}+3{{H}_{2}}O$
${{n}_{C{{O}_{2}}}}=0,672:22,4=0,03(mol)={{n}_{C{{O}_{3}}^{2-}}}$
${{n}_{HCl}}=2\times {{n}_{C{{O}_{2}}}}=2\times 0,03=0,06(mol)={{n}_{C{{l}^{-}}}}$
$\begin{align}
& {{m}_{mcacbonat}}={{m}_{KL}}+{{m}_{C{{O}_{3}}^{2-}}} \\
& \Rightarrow {{m}_{KL}}={{m}_{mcacbonat}}-{{m}_{C{{O}_{3}}^{2-}}}=10,05-60\times 0,03=8,25g \\
\end{align}$
$m={{m}_{KL}}+{{m}_{C{{l}^{-}}}}=8,25+35,5\times 0,06=10,38g$
b)
$MC{{O}_{3}}\xrightarrow{t{}^\circ }MO+C{{O}_{2}}$
${{B}_{2}}{{(C{{O}_{3}})}_{3}}\xrightarrow{t{}^\circ }{{B}_{2}}{{O}_{3}}+3C{{O}_{2}}$
Bảo toàn khối lượng ta có:
$\begin{align}
& {{m}_{mcacbonat}}=a+{{m}_{C{{O}_{2}}}} \\
& a={{m}_{mcacbonat}}-{{m}_{C{{O}_{2}}}}=10,05-0,03\times 44=8,73g \\
\end{align}$