Đáp án:
\(\begin{array}{l}
C{\% _{MgC{l_2}}} = 4,65\% \\
C{\% _{CaC{l_2}}} = 2,72\% \\
C{\% _{HC{l_d}}} = 1,788\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
X + 2HCl \to XC{l_2} + {H_2}\\
{n_{{H_2}}} = \dfrac{m}{M} = \dfrac{{6,72}}{{22,4}} = 0,3mol\\
{n_X} = {n_{{H_2}}} = 0,3mol\\
{M_X} = \dfrac{m}{n} = \dfrac{{8,8}}{{0,3}} = 29,33dvC\\
\Rightarrow A:Magie(Mg)\\
B:Canxi(Ca)\\
hh:Mg(a\,mol),Ca(b\,mol)\\
\left\{ \begin{array}{l}
a + b = 0,3\\
24a + 40b = 8,8
\end{array} \right.\\
\Rightarrow a = 0,2;b = 0,1\\
{m_{HCl}} = \dfrac{{400 \times 7,3}}{{100}} = 29,2\\
{n_{HCl}} = \dfrac{m}{M} = \dfrac{{29,2}}{{36,5}} = 0,8mol\\
\dfrac{{{n_{HCl}}}}{2} > {n_{{H_2}}} \Rightarrow HCl\text{ dư}\\
{m_{ddspu}} = 8,8 + 400 - 0,3 \times 2 = 408,2g\\
{n_{MgC{l_2}}} = {n_{Mg}} = 0,2mol\\
{m_{MgC{l_2}}} = 0,2 \times 95 = 19g\\
{n_{CaC{l_2}}} = {n_{Ca}} = 0,1mol\\
{m_{CaC{l_2}}} = 0,1 \times 111 = 11,1g\\
{n_{HC{l_d}}} = {n_{HCl}} - 2{n_{{H_2}}} = 0,2mol\\
{m_{HC{l_d}}} = 0,2 \times 36,5 = 7,3g\\
C{\% _{MgC{l_2}}} = \dfrac{{19}}{{408,2}} \times 100\% = 4,65\% \\
C{\% _{CaC{l_2}}} = \dfrac{{11,1}}{{408,2}} \times 100\% = 2,72\% \\
C{\% _{HC{l_d}}} = \dfrac{{7,3}}{{408,2}} \times 100\% = 1,788\%
\end{array}\)
