a, Ta có: 100nCaCO3 + 84nMgCO3 = 4,68 (1)
PT: \(CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\)
\(MgCO_3+2HCl\rightarrow MgCl_2+CO_2+H_2O\)
Theo PT: \(n_{CO_2}=n_{CaCO_3}+n_{MgCO_3}=\dfrac{1,2395}{24,79}=0,05\left(mol\right)\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{CaCO_3}=0,03\left(mol\right)\\n_{MgCO_3}=0,02\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{CaCO_3}=0,03.100=3\left(g\right)\\m_{MgCO_3}=0,02.84=1,68\left(g\right)\end{matrix}\right.\)
b, Theo PT: \(n_{HCl}=2n_{H_2}=0,1\left(mol\right)\Rightarrow m_{ddHCl}=\dfrac{0,1.36,5}{20\%}=18,25\left(g\right)\)
⇒ m dd sau pư = 4,68 + 18,25 - 0,05.44 = 20,73 (g)
Có: \(\left\{{}\begin{matrix}n_{CaCl_2}=n_{CaCO_3}=0,03\left(mol\right)\\n_{MgCl_2}=n_{MgCO_3}=0,02\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{CaCl_2}=\dfrac{0,03.111}{20,73}.100\%\approx16,1\%\\C\%_{MgCl_2}=\dfrac{0,02.95}{20,73}.100\%\approx9,2\%\end{matrix}\right.\)
c, \(C_{M_{HCl}}=\dfrac{0,1}{0,25}=0,4\left(M\right)\)
