Đáp án:
\(C{\% _{NaCl}} = 19,5\% \)
\(C{\% _{HCl}} = 36,5\% \)
Giải thích các bước giải:
Phản ứng xảy ra:
\(NaOH + HCl\xrightarrow{{}}NaCl + {H_2}O\)
Ta có:
\({m_{NaOH}} = 200.20\% = 40{\text{ gam}}\)
\( \to {n_{NaOH}} = \frac{{40}}{{40}} = 1{\text{ mol = }}{{\text{n}}_{NaCl}} = {n_{HCl}}\)
\( \to {m_{NaCl}} = 1.(23 + 35,5) = 58,5{\text{ gam}}\)
BTKL:
\({m_{dd}} = {m_{dd{\text{ NaOH}}}} + {m_{dd\;{\text{HCl}}}} = 200 + 100 = 300{\text{ g}}am\)
\( \to C{\% _{NaCl}} = \frac{{58,5}}{{300}}.100\% = 19,5\% \)
\({m_{HCl}} = 1.36,5 = 36,5{\text{ gam}}\)
\( \to C{\% _{HCl}} = \frac{{36,5}}{{100}}.100\% = 36,5\% \)