Đáp án:
m=4,03 gam
Giải thích các bước giải:
Phản ứng xảy ra:
\(N{a_2}O + {H_2}O\xrightarrow{{}}2NaOH\)
\(2NaOH + {H_2}S{O_4}\xrightarrow{{}}N{a_2}S{O_4} + 2{H_2}O\)
Ta có:
\({n_{{H_2}S{O_4}}} = 0,25.1 = 0,25{\text{ mol;}}{{\text{V}}_{dd\;{\text{B}}}} = 300 + 250 = 550ml = 0,55{\text{ lít}}\)
\( \to {{\text{n}}_{{H_2}S{O_4}{\text{dư}}}} = 0,55.0,2 = 0,11{\text{ mol}} \to {{\text{n}}_{{H_2}S{O_4}{\text{ phản ứng}}}} = 0,25 - 0,11 = 0,14{\text{ mol}}\)
\( \to {n_{NaOH}} = 2{n_{{H_2}S{O_4}}} = 0,14.2 = 0,28{\text{ mol}} \)
\(\to {{\text{C}}_{M{\text{ NaOH}}}} = \frac{{0,28}}{{0,3}} = 0,933M\)
\({n_{NaOH{\text{ ban đầu}}}} = 0,3.0,5 = 0,15{\text{ mol}} \to {{\text{n}}_{N{a_2}O}} = \frac{{0,28 - 0,15}}{2} = 0,065{\text{ mol}}\)
\( \to {\text{m = 0}}{\text{,065}}{\text{.62 = 4}}{\text{,03 gam}}\)
\({n_{N{a_2}S{O_4}}} = \frac{1}{2}{n_{NaOH}} = 0,14{\text{ mol}} \to {{\text{C}}_{M{\text{N}}{{\text{a}}_2}S{O_4}}} = \frac{{0,14}}{{0,55}} = 0,254M\)
