Đáp án:

\({\text{C}}{{\text{\% }}_{{H_2}S{O_4}}} = 49\% \)

Giải thích các bước giải:

 Phản ứng xảy ra:

\(BaC{l_2} + {H_2}S{O_4}\xrightarrow{{}}BaC{l_2} +2HCl\)

Trung hòa dung dịch sau phản ứng:

\(HCl + NaOH\xrightarrow{{}}NaCl + {H_2}O\)

\({H_2}S{O_4} + 2NaOH\xrightarrow{{}}N{a_2}S{O_4} + 2{H_2}O\)

\({m_{dd{\text{ NaOH}}}} = 250.1,28 = 320{\text{ gam}} \to {{\text{m}}_{NaOH}} = 320.25\%  = 80{\text{ gam}} \to {{\text{n}}_{NaOH}} = \frac{{80}}{{40}} = 2mol\)

\({n_{NaOH}} = {n_{HCl}} + 2{n_{{H_2}S{O_4}{\text{ du}}}} = 2{n_{{H_2}S{O_4}{\text{ phản ứng }}}} + 2{n_{{H_2}S{O_4}{\text{ dư}}}} = 2{n_{{H_2}S{O_4}}} \to {n_{{H_2}S{O_4}}} = 1{\text{ mol}}\)

\( \to {m_{{H_2}S{O_4}}} = 1.98 = 98{\text{ gam}} \to {\text{C}}{{\text{\% }}_{{H_2}S{O_4}}} = \frac{{98}}{{200}} = 49\% \)