Đáp án:
\({V_{{H_2}}} = 1,344l\)
Giải thích các bước giải:
\(\begin{array}{l}
{H_2}S{O_4} + 2NaOH \to N{a_2}S{O_4} + 2{H_2}O\\
{n_{{H_2}S{O_4}}} = 0,11mol\\
{n_{NaOH}} = 0,1mol\\
\to {n_{{H_2}S{O_4}}} > \dfrac{{{n_{NaOH}}}}{2}
\end{array}\)
Suy ra \({H_2}S{O_4}\) dư
\( \to {n_{{H_2}S{O_4}}}dư= 0,11 - \dfrac{1}{2}{n_{NaOH}} = 0,06mol\)
\(\begin{array}{l}
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}\\
{n_{Al}} = 0,05mol\\
\to \dfrac{{{n_{Al}}}}{2} > \dfrac{{{n_{{H_2}S{O_4}}}dư}}{3}
\end{array}\)
Suy ra Al dư
\(\begin{array}{l}
\to {n_{{H_2}}} = {n_{{H_2}S{O_4}}}dư= 0,06mol\\
\to {V_{{H_2}}} = 1,344l
\end{array}\)
