Đáp án:
Giải thích các bước giải:
a)
Ta có PTHH
$\begin{array}{l}
Mg + \,2HCl \to MgC{l_2} + {H_2}\\
a\,\,\,\,\,\,\,\,\,\,\,\,\,2a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a\,\,\,\,\,\,\,\,\,\,\,\,\,\,a\\
2Al\, + \,6HCl\, \to 2AlC{l_3} + 3{H_2}\\
b\,\,\,\,\,\,\,\,\,\,\,\,\,\,3b\,\,\,\,\,\,\,\,\,\,\,\,\,\,b\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1,5b\,\,
\end{array}$
${n_{{H_2}}} = 0,45$
Ta có hệ phương trình
$\begin{array}{l}
24a + 27b = 9\\
a + 1,5b = 0,45\,\,\,\,\,\,\, \Rightarrow a = 0,15\,\,;\,\,b = 0,2
\end{array}$
b) Các chất tan có trong X gồm $HC{l_{du\,}},\,MgC{l_2},AlC{l_3}$
\[\begin{array}{l}
{n_{HClpu}} = 2a + 3b = 0,9\\
{n_{HC{l_{du\,}}}} = 0,1.0,9 = 0,09\,\,\, \Rightarrow {n_{HClbd}} = 0,99\\
{m_{HClbd}} = 0,99.36,5 = 36,135g\\
{m_{{\rm{dd}}}}_{HCl} = \frac{{{m_{HClbd}}}}{{C\% }}.100 = 198g\\
{m_{{\rm{dd}}sau\,pu}} = {m_{kl}} + {m_{{\rm{dd}}}}_{HCl} - {m_{{H_2}}} = 9 + 198 - 0,45.2 = 206,1g
\end{array}\]
\[\begin{array}{l}
C{\% _{HCl\,du}} = \frac{{{m_{HCl\,du}}}}{{{m_{{\rm{dd}}sau\,pu}}}}.100 = \frac{{0,09.36,5.100}}{{206,1}} = 1,59\% \\
C{\% _{MgC{l_2}}} = \frac{{0,15.95}}{{206,1}}.100 = 6,91\% \\
C{\% _{AlC{l_3}}} = \frac{{0,2.133,5}}{{206,1}}.100 = 12,95\%
\end{array}\]
d) \[\begin{array}{l}
NaOH + HCl \to NaCl + H2O\\
0,09\,\,\,\, \leftarrow 0,09\\
2NaOH + MgC{l_2} \to Mg{(OH)_2} + 2NaCl\\
0,3\,\,\,\,\,\,\,\,\,\,\, \leftarrow 0,15\\
3NaOH\,\, + AlC{l_3} \to 3NaCl + Al{(OH)_3}\\
0,6\,\,\,\,\,\,\,\,\,\,\,\, \leftarrow 0,2
\end{array}\]
=> \[{n_{NaOH}} = 0,99 \to {V_{{\rm{dd}}}} = \frac{{0,99}}{2} = 0,495l\]
