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) \[\begin{array}{l} NaOH + HCl \to NaCl + H2O\\ 0,09\,\,\,\, \leftarrow 0,09\\ 2NaOH + MgC{l_2} \to Mg{(OH)_2} + 2NaCl\\ 0,3\,\,\,\,\,\,\,\,\,\,\, \leftarrow 0,15\\ 3NaOH\,\, + AlC{l_3} \to 3NaCl + Al{(OH)_3}\\ 0,6\,\,\,\,\,\,\,\,\,\,\,\, \leftarrow 0,2 \end{array}\] => \[{n_{NaOH}} = 0,99 \to {V_{{\rm{dd}}}} = \frac{{0,99}}{2} = 0,495l\]