Đáp án:
\({m_{dd{\text{ }}{{\text{H}}_2}S{O_4}}} = 50{\text{ gam}}\)
\({V_{dd{\text{HCl}}}} = 0,125{\text{ lít}}\)
Giải thích các bước giải:
\({m_{NaOH}} = 200.5\% = 10{\text{ gam}} \to {{\text{n}}_{NaOH}} = \frac{{10}}{{40}} = 0,25{\text{ mol}}\)
\(2NaOH + {H_2}S{O_4}\xrightarrow{{}}N{a_2}S{O_4} + 2{H_2}O\)
Ta có: \({n_{{H_2}S{O_4}}} = \frac{1}{2}{n_{NaOH}} = 0,125{\text{ mol}}\)
\( \to {m_{{H_2}S{O_4}}} = 0,125.98 = 12,25{\text{ gam}} \to {{\text{m}}_{dd\;{{\text{H}}_2}S{O_4}}} = \frac{{12,25}}{{24,5\% }} = 50{\text{ gam}}\)
\(NaOH + HCl\xrightarrow{{}}NaCl + {H_2}O\)
\( \to {n_{HCl}} = {n_{NaOH}} = 0,25{\text{ mol}} \to {{\text{V}}_{HCl}} = \frac{{{n_{HCl}}}}{{{C_{M{\text{ HCl}}}}}} = \frac{{0,25}}{2} = 0,125{\text{ lít}}\)
