Đáp án:
\({V_{{H_2}S{O_4}}} = 32,367{\text{ ml}}\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(N{a_2}O + {H_2}O\xrightarrow{{}}2NaOH\)
Ta có:
\({n_{N{a_2}O}} = \frac{{9,3}}{{23.2 + 16}} = 0,15{\text{ mol}}\)
\( \to {n_{NaOH}} = 2{n_{N{a_2}O}} = 0,3{\text{ mol}}\)
\( \to {C_{M{\text{ NaOH}}}} = \frac{{0,3}}{{0,5}} = 0,6M\)
500 ml \(A\) chứa 0,3 mol \(NaOH\).
Vậy 250 ml \(A\) chứa 0,15 mol \(NaOH\).
\(2NaOH + {H_2}S{O_4}\xrightarrow{{}}N{a_2}S{O_4} + 2{H_2}O\)
Ta có:
\({n_{{H_2}S{O_4}}} = \frac{1}{2}{n_{NaOH}} = 0,075{\text{ mol}}\)
\({m_{{H_2}S{O_4}}} = 0,075.98 = 7,35{\text{ gam}}\)
\( \to {m_{dd\;{{\text{H}}_2}S{O_4}}} = \frac{{7,35}}{{20\% }} = 36,75{\text{ gam}}\)
\( \to {V_{{H_2}S{O_4}}} = \frac{{36,75}}{{1,14}} = 32,367{\text{ ml}}\)
