Gọi CTHH của muối cacbonat kim loại R là $R_2(CO_3)_n$
$MgCO_3 + 2HCl \to MgCl_2 + CO_2 + H_2O$
$R_2(CO_3)_n + 2nHCl \to 2RCl_n + nCO_2 + nH_2O$
Theo PTHH : $n_{HCl} = 2n_{CO_2} = 0,3(mol)$
$m_{dd\ HCl} = \dfrac{0,3.36,5}{7,3\%} = 150(gam)$
$m_{dd\ B} = m_A + m_{dd\ HCl} - m_{CO_2} = 14,2 + 150 - 0,15.44 = 157,6(gam)$
$\Rightarrow n_{MgCl_2} = \dfrac{157,6.6,028\%}{95} = 0,1(mol)$
Theo PTHH :
$n_{CO_2} = n.n_{R_2(CO_3)_n} + n_{MgCO_3} = 0,15$
$\Rightarrow n_{R_2(CO_3)_n} = \dfrac{0,05}{n}(mol)$
$\Rightarrow 0,1.84 + \dfrac{0,05}{n}(2R + 60n) = 14,2$
$\Rightarrow R = 28n$
Với n = 2 thì R = 56(Fe)
$\%m_{MgCO_3} = \dfrac{0,1.84}{14,2}.100\% = 59,15\%$
$\%m_{FeCO_3} = 100\% - 59,15\% = 40,85\%$
Gọi hoá trị của R là n
Ta có: \(n_{CO_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
PTHH: \(MgCO_3+2HCl\rightarrow MgCl_2+CO_2+H_2O\) (1)
\(R_2\left(CO_3\right)_n+2nHCl\rightarrow2RCl_n+nCO_2+nH_2O\) (2)
Theo PT (1), (2): \(n_{HCl}=2n_{CO_2}=0,3\left(mol\right)\)
=> \(m_{ddHCl}=\dfrac{0,3.36,5}{7,3\%}=150\left(g\right)\)
=> \(m_{ddspư}=150+14,2-0,15.44=157,6\left(g\right)\)
=> \(m_{MgCl_2}=157,6.6,028\%=9,5\left(g\right)\)
=> \(n_{MgCl_2}=\dfrac{9,5}{95}=0,1\left(mol\right)\)
Theo PT (1): \(n_{MgCO_3}=n_{MgCl_2}=0,1\left(mol\right)\)
=> \(m_{R_2\left(CO_3\right)_n}=14,2-0,1.84=5,8\left(g\right)\)
Theo PT (1), (2): \(n_{CO_2}=n_{MgCO_3}+n.n_{R_2\left(CO_3\right)_n}\)
=> \(n_{R_2\left(CO_3\right)_n}=\dfrac{0,15-0,1}{n}=\dfrac{0,05}{n}\left(mol\right)\)
=> \(M_{R_2\left(CO_3\right)_n}=\dfrac{5,8}{\dfrac{0,05}{n}}=116n\left(g/mol\right)\)
=> \(M_R=\dfrac{116n-60n}{2}=28n\left(g/mol\right)\)
Xét n = 2 t/m => MR = 56 (g/mol)
=> R là Fe
=> \(\left\{{}\begin{matrix}\%m_{MgCO_3}=\dfrac{0,1.84}{14,2}.100\%=59,15\%\\\%m_{FeCO_3}=100\%-59,15\%=40,85\%\end{matrix}\right.\)