Đáp án:
\(C{\% _{NaOH}} = 32\% \)
\(C{\% _{N{a_2}S{O_4}}} = 18,933\% \)
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:
\({m_{{H_2}S{O_4}}} = 50.19,6\% = 9,8{\text{ gam}}\)
\( \to {n_{{H_2}S{O_4}}} = \frac{{9,8}}{{98}} = 0,1{\text{ mol}}\)
\( \to {n_{NaOH}} = 2{n_{{H_2}S{O_4}}} = 0,2{\text{ mol}}\)
\( \to {m_{NaOH}} = 0,2.40 = 8{\text{ gam}}\)
\( \to C{\% _{NaOH}} = \frac{8}{{25}} = 32\% \)
BTKL:
\({m_{dd{\text{ sau phản ứng}}}} = 50 + 25 = 75{\text{ gam}}\)
\({n_{N{a_2}S{O_4}}} = {n_{{H_2}S{O_4}}} = 0,1{\text{ mol}}\)
\( \to {m_{N{a_2}S{O_4}}} = 0,1.(23.2 + 32 + 16.4) = 14,2{\text{ gam}}\)
\( \to C{\% _{N{a_2}S{O_4}}} = \frac{{14,2}}{{75}} = 18,933\% \)
