Đáp án:
\( {m_{dd{\text{ C}}{{\text{H}}_3}COOH}} = 901,5624{\text{ gam}}\)
Giải thích các bước giải:
Phản ứng xảy ra:
\({C_2}{H_5}OH + {O_2}\xrightarrow{{men}}C{H_3}COOH + {H_2}O\)
\({V_{{C_2}{H_5}OH}} = 50.96\% = 48{\text{ ml}}\)
\( \to {m_{{C_2}{H_5}OH}} = 48.0,8 = 38,4{\text{ gam}}\)
\( \to {n_{{C_2}{H_5}OH}} = \frac{{38,4}}{{46}} = 0,83478{\text{ mol = }}{{\text{n}}_{C{H_3}COOH{\text{ lt}}}}\)
\( \to {m_{C{H_3}COOH}} = 0,83478.90\% = 0,751302{\text{ mol}}\)
\( \to {m_{C{H_3}COOH}} = 0,751302.60 = 45,07812{\text{ gam}}\)
\( \to {m_{dd{\text{ C}}{{\text{H}}_3}COOH}} = \frac{{45,07812}}{{5\% }} = 901,5624{\text{ gam}}\)
