Đáp án:
\(C{\% _{N{a_2}C{O_3}}} = 53\% ;C{\% _{C{H_3}COOH}} = 30\% \)
Giải thích các bước giải:
Phản ứng xảy ra:
\(2C{H_3}COOH + N{a_2}C{O_3}\xrightarrow{{}}2C{H_3}COONa + C{O_2} + {H_2}O\)
Ta có:
\({n_{C{O_2}}} = \frac{{11,2}}{{22,4}} = 0,5{\text{ mol = }}{{\text{n}}_{N{a_2}C{O_3}}} \to {n_{C{H_3}COOH}} = 2{n_{C{O_2}}} = 1{\text{ mol}}\)
\(\to {m_{N{a_2}C{O_3}}} = 0,5.106 = 53{\text{ gam;}}{{\text{m}}_{C{H_3}COOH}} = 1.60 = 60{\text{ gam}}\)
\( \to C{\% _{N{a_2}C{O_3}}} = \frac{{53}}{{100}} = 53\% ;C{\% _{C{H_3}COOH}} = \frac{{60}}{{200}} = 30\% \)
BTKL: \({m_{dd{\text{ sau phản ứng}}}} = 100 + 200 - 0,5.44 = 278{\text{ gam}}\)
\({n_{C{H_3}COONa}} = {n_{C{H_3}COOH}} = 1{\text{ mol}} \to {{\text{m}}_{C{H_3}COONa}} = 1.82 = 82{\text{ gam}} \to C{\% _{C{H_3}COONa}} = \frac{{82}}{{278}} = 29,5\% \)
