Giải thích các bước giải:
\(\begin{array}{l}
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
a)\\
{n_{HCl}} = 0,075mol\\
{n_{{H_2}}} = 0,03mol\\
\to {n_{HCl}} > {n_{{H_2}}} \to {n_{HCl}}dư\\
\to {n_{Al}} = {n_{AlC{l_3}}} = \dfrac{2}{3}{n_{{H_2}}} = 0,02mol\\
\to {m_{Al}} = 0,54g\\
\to {m_S} = 1,18 - 0,54 = 0,64g \to {n_S} = 0,02mol\\
\to {n_{HCl(pt)}} = 2{n_{{H_2}}} = 0,06mol\\
\to {n_{HCl(dư)}} = 0,015mol\\
\to C{M_{HCl(dư)}} = \dfrac{{0,015}}{{0,375}} = 0,04M\\
\to C{M_{AlC{l_3}}} = \dfrac{{0,02}}{{0,375}} = 0,053M\\
b)\\
2Al + 3S \to A{l_2}{S_3}\\
{m_{hỗnhợpA}}(b) = 3{m_{hỗnhợpA}}(a)\\
\to {n_{Al}} = 3{n_{Al}}(a) = 0,06mol\\
\to {n_S} = 3{n_S}(a) = 0,06mol\\
\to \frac{{{n_{Al}}}}{2} > \frac{{{n_S}}}{3} \to {n_{Al}}dư\\
{n_{Al(pt)}} = \dfrac{2}{3}{n_S} = 0,04mol\\
\to {n_{Al(dư)}} = 0,02mol\\
\to {m_{Al(dư)}} = 0,54g\\
{n_{A{l_2}{S_3}}} = \dfrac{1}{3}{n_S} = 0,02mol \to {m_{A{l_2}{S_3}}} = 3g\\
\to \% {m_{A{l_2}{S_3}}} = \dfrac{3}{{3 + 0,54}} \times 100\% = 85\% \\
\to \% {m_{Al(dư)}} = \dfrac{{0,54}}{{3 + 0,54}} \times 100\% = 15\%
\end{array}\)