Đáp án:
\({\text{C}}{{\text{\% }}_{C{H_3}COOH}} = 4\% \)
a%=3,92%
Giải thích các bước giải:
Phản ứng xảy ra:
\(C{H_3}COOH + KOH\xrightarrow{{}}C{H_3}COOK + {H_2}O\)
Ta có:
\({m_{KOH}} = 200.5,6\% = 11,2{\text{ gam}} \to {{\text{n}}_{KOH}} = {n_{C{H_3}COOH}} = {n_{C{H_3}COOK}} = \frac{{11,2}}{{56}} = 0,2{\text{ mol}}\)
Ta có:
\({m_{C{H_3}COOH}} = 0,2.60 = 12{\text{ gam}} \to {\text{C}}{{\text{\% }}_{C{H_3}COOH}} = \frac{{12}}{{300}} = 4\% \)
Ta có:
\({m_{dd}} = 300 + 200 = 500{\text{ gam;}}{{\text{m}}_{C{H_3}COOK}} = 0,2.(12 + 3 + 44 + 39) = 19,6{\text{ gam}} \to {\text{a\% = }}\frac{{19,6}}{{500}} = 3,92\% \)
