Đáp án:
\( {m_{dd\;{\text{N}}{{\text{a}}_2}C{O_3}}} = 200{\text{ gam}}\)
\( C{\% _{C{H_3}COONa}} = 21,69\% \)
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:
\({m_{C{H_3}COOH}} = 200.30\% = 60{\text{ gam}}\)
\( \to {n_{C{H_3}COOH}} = \frac{{60}}{{60}} = 1{\text{ mol}}\)
\( \to {n_{N{a_2}C{O_3}}} = \frac{1}{2}{n_{C{H_3}COOH}} = 0,5{\text{ mol}}\)
\( \to {m_{N{a_2}C{O_3}}} = 0,5.(23.2 + 12 + 16.3) = 53{\text{ gam}}\)
\( \to {m_{dd\;{\text{N}}{{\text{a}}_2}C{O_3}}} = \frac{{53}}{{26,5\% }} = 200{\text{ gam}}\)
\({n_{C{H_3}COONa}} = {n_{C{H_3}COOH}} = 1{\text{ mol;}}{{\text{n}}_{C{O_2}}} = \frac{1}{2}{n_{C{H_3}COOH}} = 0,5{\text{ mol}}\)
BTKL:
\({m_{dd{\text{ C}}{{\text{H}}_3}COOH}} + {n_{dd\;{\text{N}}{{\text{a}}_2}C{O_3}}} = {m_{dd}} + {m_{C{O_2}}}\)
\( \to 200 + 200 = {m_{dd}} + 0,5.44 \to {m_{dd}} = 378{\text{ gam}}\)
\( \to {m_{C{H_3}COONa}} = 1.82 = 82{\text{ gam}}\)
\( \to C{\% _{C{H_3}COONa}} = \frac{{82}}{{378}} = 21,69\% \)
