Đáp án:
\( {m_{NaOH{\text{ dư}}}} = 25,6{\text{ gam}}\)
\( {C_{M{\text{ NaOH}}}} = 1,73M;{C_{M{\text{ N}}{{\text{a}}_2}S{O_4}}} = 0,081M\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(2NaOH + {H_2}S{O_4}\xrightarrow{{}}N{a_2}S{O_4} + 2{H_2}O\)
Ta có:
\({n_{NaOH}} = 0,35.2 = 0,7{\text{ mol}}\)
\({n_{{H_2}S{O_4}}} = 0,02.1,5 = 0,03{\text{ mol < }}\frac{1}{2}{n_{NaOH}}\)
Vậy \(NaOH\) dư
\( \to {n_{NaOH{\text{ dư}}}} = 0,7 - 0,03.2 = 0,64{\text{ mol}}\)
\( \to {m_{NaOH{\text{ dư}}}} = 0,64.40 = 25,6{\text{ gam}}\)
\({n_{N{a_2}S{O_4}}} = {n_{{H_2}S{O_4}}} = 0,03{\text{ mol}}\)
\({V_{dd}} = 350 + 20 = 370{\text{ ml = 0}}{\text{,37 lít}}\)
\( \to {C_{M{\text{ NaOH}}}} = \frac{{0,64}}{{0,37}} = 1,73M;{C_{M{\text{ N}}{{\text{a}}_2}S{O_4}}} = \frac{{0,03}}{{0,37}} = 0,081M\)