$\begin{align}
& 2R{{(HC{{O}_{3}})}_{n}}\to {{R}_{2}}{{(C{{O}_{3}})}_{n}}+nC{{O}_{2}}+n{{H}_{2}}O \\
& \,\,\,\,\,\,\,\,2a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a\times n\,\,\,\,\,\,\,\,\,a\times n \\
\end{align}$ (1)
Có thể xảy ra phản ứng:
$\begin{align}
& {{R}_{2}}{{(C{{O}_{3}})}_{n}}\to {{R}_{2}}{{O}_{n}}+nC{{O}_{2}} \\
& \,\,\,\,\,a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a\times n \\
\end{align}$ (2)
${{n}_{\downarrow }}={{n}_{CaC{{O}_{3}}}}=4:100=0,04(mol)$
Ta xét hai trường hợp.
+ Trường hợp 1 : $Ca{{(OH)}_{2}}$ dư, chỉ xảy ra phản ứng
$C{{O}_{2}}+Ca{{(OH)}_{2}}\to CaC{{O}_{3}}+{{H}_{2}}O$
${{n}_{C{{O}_{2}}}}={{n}_{CaC{{O}_{3}}}}=0,04(mol)$
${{m}_{binh\,\,\tan g}}={{m}_{C{{O}_{2}}}}+{{m}_{{{H}_{2}}O}}=44\times 0,04+18\times {{n}_{{{H}_{2}}O}}=5,3$
$\Rightarrow {{n}_{{{H}_{2}}O}}=0,197(mol)>{{n}_{C{{O}_{2}}}}$ (loại, không thỏa mãn PTHH (1) và (2))
+ Trường hợp 2: $C{{O}_{2}}$ dư , xảy ra 2 phản ứng
$\begin{align}
& C{{O}_{2}}+Ca{{(OH)}_{2}}\to CaC{{O}_{3}}+{{H}_{2}}O \\
& 0,07\leftarrow 0,07\to \,\,\,\,\,\,\,\,\,\,\,\,0,07 \\
\end{align}$
$C{{O}_{2}}+CaC{{O}_{3}}+{{H}_{2}}O\to Ca{{(HC{{O}_{3}})}_{2}}$ (*)
${{n}_{CaC{{O}_{3}}}}=0,04(mol)\Rightarrow {{n}_{CaC{{O}_{3}}}}_{(*)}=0,07-0,04=0,03(mol)={{n}_{C{{O}_{2}}(*)}}$
${{n}_{C{{O}_{2}}}}=0,07+0,03=0,1(mol)$
${{m}_{binh\,\,\tan g}}={{m}_{C{{O}_{2}}}}+{{m}_{{{H}_{2}}O}}=44\times 0,1+18\times {{n}_{{{H}_{2}}O}}=5,3$
$\Rightarrow {{n}_{{{H}_{2}}O}}=0,05(mol)$
${{n}_{C{{O}_{2}}}}=2\times {{n}_{{{H}_{2}}O}}\Rightarrow $xảy ra phản ứng (2)
Ta có : $a\times n=0,05\Rightarrow a=\frac{0,05}{n}$
${{m}_{R{{(HC{{O}_{3}})}_{n}}}}=2a\times (R+61\times n)=2\times \frac{0,05}{n}\times (R+61\times n)=12,95$
$\begin{align}
& \Rightarrow 0,1\times R+0,1\times 61\times n=12,95\times n \\
& \Rightarrow R=68,5\times n \\
\end{align}$
Vì R là kim loại nên n chỉ có thể là : 1, 2 hoặc 3.
Khi n = 2 thì R = 137 (Ba)
CT muối là: $Ba{{(HC{{O}_{3}})}_{2}}$