Đáp án:
\({\text{C}}{{\text{\% }}_{C{H_3}COOK}} = 11,9\% \)
Giải thích các bước giải:
Phản ứng xảy ra:
\(2C{H_3}COOH + {K_2}C{O_3}\xrightarrow{{}}2C{H_3}COOK + C{O_2} + {H_2}O\)
Ta có:
\({m_{C{H_3}COOH}} = 200.12\% = 24{\text{ gam}} \to {{\text{n}}_{C{H_3}COOH}} = \frac{{24}}{{60}} = 0,4{\text{ mol}} \to {{\text{n}}_{{K_2}C{O_3}}} = {n_{C{O_2}}} = \frac{1}{2}{n_{C{H_3}COOH}} = 0,2{\text{ mol}}\)
\( \to {V_{C{O_2}}} = 0,2.22,4 = 4,48{\text{ lít}}\)
Ta có:
\({m_{{K_2}C{O_3}}} = 0,2.(39.2 + 60) = 27,6{\text{gam}} \to {{\text{m}}_{dd\;{{\text{K}}_2}C{O_3}}} = \frac{{27,6}}{{20\% }} = 138{\text{ gam}}\)
BTKL:
\({m_{dd{\text{sau phản ứng}}}} = 200 + 138 - 0,2.44 = 329,2{\text{ gam}}\)
\({n_{C{H_3}COOK}} = {n_{C{H_3}COOH}} = 0,4{\text{ mol}} \to {{\text{m}}_{C{H_3}COOK}} = 0,4.(60 + 38) = 39,2{\text{ gam}} \to {\text{C}}{{\text{\% }}_{C{H_3}COOK}} = \frac{{39,2}}{{329,2}} = 11,9\% \)
