Giải thích các bước giải:

$m_{dd\ HCl}=100×1,047=104,7(g)$

$n_{HCl}=\dfrac{104,7×19,87\%}{36,5}=0,57(mol)$

$2HCl+Na_2CO_3\to 2NaCl+CO_2+H_2O$

$n_{CO_2}=\dfrac{1,904}{22,4}=0,085(mol)$

$n_{HCl\ \rm (dư)}=2n_{CO_2}=0,17(mol)$

$n_{HCl\ \rm (phản\ ứng)}=0,57-0,17=0,4(mol)$

Gọi: $\begin{cases} n_{MgO}=x(mol)\\n_{CaO}=y(mol) \end{cases}$

$\to 40x+56y=9,6(g)\ (1)$

$MgO+2HCl\to MgCl_2+H_2O$

$CaO+2HCl\to CaCl_2+H_2O$

$n_{HCl}=2x+2y=0,4(mol)\ (2)$

Từ $(1),(2),$ ta giải được:

$\begin{cases} x=0,1(mol)\\y=0,1(mol) \end{cases}$

$\%m_{MgO}=\dfrac{0,1×40}{9,6}×100\%=41,67\%$

$\%m_{CaO}=100\%-41,67\%=58,33\%$

$n_{MgCl_2}=n_{MgO}=0,1(mol)$

$n_{CaCl_2}=n_{CaO}=0,1(mol)$

$m_{dd\ sau}=9,6+104,7=114,3(g)$

$C\%_{MgCl_2}=\dfrac{0,1×95}{114,3}×100\%=8,3\%$

$C\%_{CaCl_2}=\dfrac{0,1×111}{114,3}×100\%=9,7\%$

$C\%_{HCl\ \rm (dư)}=\dfrac{0,17×36,5}{114,3}×100\%=5,4\%$

\(m_{ddHCl}^{} = 100.1,047 = 104,7g\)

\(n_{HCl}^{} = \frac{{104,7.19,87\% }}{{36,5}} = 0,57\;{\text{mol}}\)

\(n_{CO_2^{}}^{} = \frac{{1,904}}{{22,4}} = 0,085\;mol\)

\(Na_2^{}CO_3^{} + 2HCl \to 2NaCl + CO_2^{} + H_2^{}O\)

\({\text{                 0}}{\text{,17                      0}}{\text{,085}}\)

\({\text{n}}_{HCl_{PU}^{}}^{} = 0,57 - 0,17 = 0,4\;{\text{mol}}\)

\(Goi\;{\text{MgO : x mol; CaO :}}\;{\text{y mol}}\)

\(MgO + 2HCl \to MgCl_2^{} + H_2^{}O\)

\({\text{ x          2x}}\)

\(CaO + 2HCl \to CaCl_2^{} + H_2^{}O\)

\({\text{y           2y}}\)

\({\text{40x  +  56y  =  9}}{\text{,6}}\)

\(2x + 2y = 0,4\)

\( \to x = y = 0,1\)

\(m_{dd_{SauPU}^{}}^{} = 9,6 + 104,7 = 114,3g\)

\(C\% _{MgCl_2^{}}^{} = \frac{{0,1.95}}{{114,3}}.100 = 8,3\% \)

\(C\% _{CaCl_2^{}}^{} = \frac{{0,1.111}}{{114,3}}.100 = 9,7\% \)

\(C\% _{HCl_{Du}^{}}^{} = \frac{{0,17.36,5}}{{114,3}}.100 = 5,4\% \)

\(Na_2^{}CO_3^{} + 2HCl \to 2NaCl + CO_2^{} + H_2^{}O\)

\({\text{   0}}{\text{,085      0}}{\text{,17        0}}{\text{,17       0}}{\text{,085}}\)

\(BTKL:{\text{m}}_{dd_{sauPU}^{}}^{} + m_A^{} = {\text{m}}_{dd_{sauPU2}^{}}^{} + m_{CO_2^{}}^{}\)

\({\text{m}}_{dd_{sauPU2}^{}}^{} = 114,3 + 0,085.106 - 0,085.44\)

\({\text{m}}_{dd_{sauPU2}^{}}^{} = 119,57g\)

\(C\% _{NaCl}^{} = \frac{{0,17.58,5}}{{119,57}}.100 = 8,3\% \)