Đáp án:
\(\begin{array}{l}
a)\\
{V_{{H_2}}} = 6,72l\\
b)\\
{C_{{M_{A{l_2}{{(S{O_4})}_3}}}}} = 0,25M\\
{C_{{M_{{H_2}S{O_4}}}}} = 1,25M
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}\\
{n_{Al}} = \dfrac{m}{M} = \dfrac{{5,4}}{{27}} = 0,2mol\\
{n_{{H_2}S{O_4}}} = 0,4 \times 2 = 0,8mol\\
\dfrac{{0,2}}{2} < \dfrac{{0,8}}{3} \Rightarrow {H_2}S{O_4}\text{ dư}\\
{n_{{H_2}}} = \dfrac{3}{2}{n_{Al}} = 0,3mol\\
{V_{{H_2}}} = 0,3 \times 22,4 = 6,72l\\
b)\\
{n_{A{l_2}{{(S{O_4})}_3}}} = \dfrac{{{n_{Al}}}}{2} = 0,1mol\\
{n_{{H_2}S{O_4}d}} = {n_{{H_2}S{O_4}}} - \dfrac{3}{2}{n_{Al}} = 0,5mol\\
{C_{{M_{A{l_2}{{(S{O_4})}_3}}}}} = \dfrac{{0,1}}{{0,4}} = 0,25M\\
{C_{{M_{{H_2}S{O_4}}}}} = \dfrac{{0,5}}{{0,4}} = 1,25M
\end{array}\)
