Đáp án:
2. $ {m_{d\,d}} = 65,7\,\,gam$
3. $C{\% _{AlC{l_3}}} = 11,93\% $
Giải thích các bước giải:
1. PTHH: $2Al + 6HCl \to 2AlC{l_3} + 3{H_2}$
2. ${n_{Al}} = \dfrac{{1,62}}{{27}} = 0,06\,\,mol$
Theo PTHH: ${n_{HCl}} = 3{n_{Al}} = 3.0,06 = 0,18\,\,mol$
$ \to {m_{HCl}} = 0,18.36,5 = 6,57\,\,gam$
$ \to {m_{d\,d}} = 6,57:10\% = 65,7\,\,gam$
3. ${n_{{H_2}}} = \dfrac{3}{2}{n_{Al}} = 0,09\,\,mol$
Khối lượng dung dịch sau phản ứng:
$\begin{gathered} {m_{{\text{dd}}\,\,sau}} = {m_{Al}} + {m_{{\text{dd}}\,\,HCl}} - {m_{{H_2}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1,62 + 65,7 - 0,09.2 = 67,14\,\,gam \hfill \\ \end{gathered} $
Ta có:
$\begin{gathered} {n_{Al}} = {n_{AlC{l_3}}} = 0,06\,\,mol\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \hfill \\ \to {m_{AlC{l_3}}} = 0,06.133,5 = 8,01\,\,gam \hfill \\ \end{gathered} $
$C{\% _{AlC{l_3}}} = \dfrac{{8,01}}{{67,14}}.100\% = 11,93\% $
