Đáp án:
\(\begin{array}{l}
a){m_{Al(dư)}} = 6,75g\\
b){m_{AlC{l_3}}} = 26,7g\\
c){V_{{H_2}}} = 6,72l\\
d)C{\% _{AlC{l_3}}} = 22,06\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
a)\\
{n_{Al}} = 0,45mol\\
{m_{HCl}} = \dfrac{{109,5 \times 20\% }}{{100\% }} = 21,9g\\
\to {n_{HCl}} = 0,6mol\\
\to \dfrac{{{n_{Al}}}}{2} > \dfrac{{{n_{HCl}}}}{6} \to {n_{Al}}dư\\
\to {n_{Al(pt)}} = \dfrac{1}{3}{n_{HCl}} = 0,2mol\\
\to {n_{Al(dư)}} = 0,25mol \to {m_{Al(dư)}} = 6,75g
\end{array}\)
\(\begin{array}{l}
b)\\
{n_{AlC{l_3}}} = \dfrac{1}{3}{n_{HCl}} = 0,2mol\\
\to {m_{AlC{l_3}}} = 26,7g
\end{array}\)
\(\begin{array}{l}
c)\\
{n_{{H_2}}} = \dfrac{1}{2}{n_{HCl}} = 0,3mol\\
\to {V_{{H_2}}} = 6,72l
\end{array}\)
\(\begin{array}{l}
d)\\
{m_{{\rm{dd}}}} = {m_{Al}} + {m_{{\rm{dd}}HCl}} - {m_{{H_2}}} = 121,05g\\
\to C{\% _{AlC{l_3}}} = \dfrac{{26,7}}{{121,05}} \times 100\% = 22,06\%
\end{array}\)
