Đáp án:
\({m_{dd\;{{\text{H}}_2}S{O_4}}} = 50{\text{ gam}}\)
\(C{\% _{HCl}} = 20\% \)
Giải thích các bước giải:
Phản ứng xảy ra:
\(NaCl + {H_2}S{O_4}\xrightarrow{{{t^o}}}NaHS{O_4} + HCl\)
Ta có:
\({n_{NaCl}} = \frac{{29,25}}{{23 + 35,5}} = 0,5{\text{ mol = }}{{\text{n}}_{{H_2}S{O_4}}} = {n_{HCl}}\)
\( \to {m_{{H_2}S{O_4}}} = 0,55.98 = 49{\text{ gam}}\)
\( \to {m_{dd\;{{\text{H}}_2}S{O_4}}} = \frac{{49}}{{98\% }} = 50{\text{ gam}}\)
\({m_{HCl}} = 0,5.36,5 = 18,25{\text{ gam}}\)
Hòa tan \(HCl\) vào 73 gam nước
\( \to {m_{dd}} = 73 + 18,25 = 91,25{\text{ gam}}\)
\( \to C{\% _{HCl}} = \frac{{18,25}}{{91,25}} = 20\% \)
