Đáp án
\({V_{{H_2}}} = 4,48{\text{ lít}}\)
Giải thích:
Ta có:
\({n_{KOH}} = 0,2.1 = 0,2{\text{ mol}}\)
\({n_{{H_2}S{O_4}}} = 0,3.1 = 0,3{\text{ mol > }}\frac{1}{2}{n_{KOH}}\)
Vậy axit dư.
Các phản ứng xảy ra:
\(2KOH + {H_2}S{O_4}\xrightarrow{{}}{K_2}S{O_4} + 2{H_2}O\)
\({H_2}S{O_4} + Fe\xrightarrow{{}}FeS{O_4} + {H_2}\)
Ta có:
\({n_{{H_2}S{O_4}{\text{ dư}}}} = 0,3 - \frac{{0,2}}{1} = 0,2{\text{ mol}}\)
\( \to {n_{{H_2}}} = {n_{{H_2}S{O_4}{\text{ dư}}}} = 0,2{\text{ mol}}\)
\( \to {V_{{H_2}}} = 0,2.22,4 = 4,48{\text{ lít}}\)
