Đáp án:
\({m_{C{H_3}COOH}} = 768{\text{ gam; }}{{\text{m}}_{dd{\text{ giấm}}}}{\text{ = 19}}{\text{,2kg}}\)
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:
\({V_{{C_2}{H_5}OH}} = 10.8\% = 0,8{\text{ lít = 800 ml}} \to {{\text{m}}_{{C_2}{H_5}OH}} = 800.0,8 = 640{\text{ gam}}\)
\({n_{{C_2}{H_5}OH}} = \frac{{640}}{{46}} = \frac{{320}}{{23}}{\text{ mol}} \to {{\text{n}}_{{C_2}{H_5}OH{\text{ phan ung}}}} = \frac{{320}}{{23}}.92\% = 12,8{\text{ mol = }}{{\text{n}}_{C{H_3}COOH}}\)
\( \to {m_{C{H_3}COOH}} = 12,8.60 = 768{\text{ gam}} \to {{\text{m}}_{dd{\text{ giấm}}}} = \frac{{768}}{{4\% }} = 19200{\text{ gam = 19}}{\text{,2kg}}\)