$2M+2nHCl\to 2MC{{l}_{n}}+n{{H}_{2}}$
Dung dịch D {HCl dư, $MC{{l}_{n}}$}
$HCl+NaHC{{O}_{3}}\to NaCl+C{{O}_{2}}+{{H}_{2}}O$
Dung dịch E {NaCl, $MC{{l}_{n}}$}
$MC{{l}_{n}}+nNaOH\to M{{(OH)}_{n}}\downarrow +nNaCl$
Nung kết tủa:
$2M{{(OH)}_{n}}\downarrow \to {{M}_{2}}{{O}_{n}}+n{{H}_{2}}O$
${{m}_{NaHC{{O}_{3}}}}=\frac{240\times 7}{100}=16,8g$
${{n}_{NaHC{{O}_{3}}}}=16,8:84=0,2mol$
$HCl+NaHC{{O}_{3}}\to NaCl+C{{O}_{2}}+{{H}_{2}}O$
$0,2\,\,\,\,\,\,\,\,\,\,0,2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,2$
${{m}_{\text{dd}E}}=\frac{0,2\times 58,5\times 100}{2,5}=468g$
${{m}_{MC{{\operatorname{l}}_{n}}}}=\frac{468\times 8,12}{100}=38g$
${{n}_{MC{{l}_{n}}}}=\frac{38}{M+n\times 35,5}$
${{n}_{{{M}_{2}}{{O}_{n}}}}=\frac{16}{2\times M+n\times 16}$
Ta có:
$\begin{align}
& \frac{38}{M+n\times 35,5}=2\times \frac{16}{2\times M+n\times 16} \\
& \frac{38}{M+n\times 35,5}=\frac{16}{M+n\times 8} \\
& 38\times (M+n\times 8)=16\times (M+n\times 35,5) \\
& 22\times M=264 \\
& M=12\times n \\
\end{align}$
Khi n = 12 thì M = 24 (Mg)
${{n}_{MgC{{l}_{2}}}}=\frac{38}{95}=0,4mol={{m}_{Mg}}$
$Mg+2HCl\to MgC{{l}_{2}}+{{H}_{2}}$
$0,4\,\,\,\,\,\,\,\,0,8\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,4$
Bảo toàn khối lượng:
$\begin{align}
& {{m}_{E}}={{m}_{Mg}}+{{m}_{\text{dd}HCl}}+{{m}_{\text{dd}NaHC{{O}_{3}}}}-{{m}_{{{H}_{2}}}}-{{m}_{C{{O}_{2}}}} \\
& =0,4\times 24+{{m}_{\text{dd}HCl}}+240-0,4\times 2-0,2\times 44=468 \\
& \Rightarrow {{m}_{\text{dd}HCl}}=228g \\
\end{align}$
${{n}_{HCl}}=0,8+0,2=1(mol)$
C% = $\frac{1\times 36,5}{228}\times 100$% = 16%