Đáp án:
\( {m_{C{H_3}COOH}} = 281,742{\text{ gam}}\)
\( {m_{giấm}} = 5634,84{\text{ ga}}m\)
Giải thích các bước giải:
Phản ứng xảy ra:
\({C_2}{H_5}OH + {O_2}\xrightarrow{{men}}C{H_3}COOH + {H_2}O\)
Ta có:
\({V_{{C_2}{H_5}OH}} = 2000.15\% = 300{\text{ ml}}\)
\( \to {m_{{C_2}{H_5}OH}} = 300.0,8 = 240{\text{ gam}}\)
\( \to {n_{{C_2}{H_5}OH}} = \frac{{240}}{{46}} = \frac{{120}}{{23}} = {n_{C{H_3}COOH{\text{ lt}}}}\)
\( \to {n_{C{H_3}COOH}} = \frac{{120}}{{23.}}90\% = 4,6957{\text{ mol}}\)
\( \to {m_{C{H_3}COOH}} = 4,6957.60 = 281,742{\text{ gam}}\)
\( \to {m_{giấm}} = \frac{{281,742}}{{5\% }} = 5634,84{\text{ ga}}m\)
