Đáp án:
\( {m_{Al}} = 5,4{\text{ gam; }}{{\text{V}}_{{H_2}}} = 6,72{\text{ lít}}\)
\({V_{dd\;{{\text{H}}_2}S{O_4}}} = 18,148{\text{ ml}}\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(2Al + 6HCl\xrightarrow{{}}2AlC{l_3} + 3{H_2}\)
Ta có:
\({m_{HCl}} = 150.14,6\% = 21,9{\text{ gam}}\)
\( \to {n_{HCl}} = \frac{{21,9}}{{36,5}} = 0,6{\text{ mol}}\)
\( \to {n_{Al}} = \frac{1}{3}{n_{HCl}} = 0,2{\text{ mol}}\)
\( \to {n_{{H_2}}} = \frac{1}{2}{n_{HCl}} = 0,3{\text{ mol}}\)
\( \to {m_{Al}} = 0,2.27 = 5,4{\text{ gam; }}{{\text{V}}_{{H_2}}} = 0,3.22,4 = 6,72{\text{ lít}}\)
Ta có:
\({n_{Al}} = \frac{{8,1}}{{27}} = 0,3{\text{ mol > }}\frac{1}{3}{n_{HCl}}\)
Vậy \(Al\) dư
\( \to {n_{Al{\text{ dư}}}} = 0,3 - \frac{1}{3}.0,6 = 0,1{\text{ mol}}\)
\(2Al + 6{H_2}S{O_4}\xrightarrow{{{t^o}}}A{l_2}{(S{O_4})_3} + 3S{O_2} + 6{H_2}O\)
\( \to {n_{{H_2}S{O_4}}} = 3{n_{Al}} = 0,3{\text{ mol}}\)
\( \to {m_{{H_2}S{O_4}}} = 0,3.98 = 29,4{\text{ gam}}\)
\( \to {m_{dd\;{{\text{H}}_2}S{O_4}}} = \frac{{29,4}}{{90\% }} = 32,667{\text{ gam}}\)
\( \to {V_{dd\;{{\text{H}}_2}S{O_4}}} = \frac{{32,667}}{{1,8}} = 18,148{\text{ ml}}\)
