Đáp án:
\(\begin{array}{l}
a)\\
{m_{AlC{l_3}}} = 8,9g\\
b)\\
{m_{Fe}} = 3,73g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
{n_{HCl}} = \dfrac{{7,3}}{{36,5}} = 0,2mol\\
{n_{Al}} = \dfrac{{5,4}}{{27}} = 0,2mol\\
\dfrac{{0,2}}{6} < \dfrac{{0,2}}{2} \Rightarrow Al\text{ dư}\\
{n_{AlC{l_3}}} = \dfrac{{{n_{HCl}}}}{3} = \dfrac{1}{{15}}mol\\
{m_{AlC{l_3}}} = \dfrac{1}{{15}} \times 133,5 = 8,9g\\
b)\\
{n_{{H_2}}} = \dfrac{{{n_{HCl}}}}{2} = 0,1mol\\
F{e_2}{O_3} + 3{H_2} \to 2Fe + 3{H_2}O\\
{n_{Fe}} = \dfrac{2}{3}{n_{{H_2}}} = \dfrac{1}{{15}}mol\\
{m_{Fe}} = \dfrac{1}{{15}} \times 56 = 3,73g
\end{array}\)
