Đáp án:
a, \({m_{C{H_3}COOH}}thực tế= \)144g
b, m=2880g
Giải thích các bước giải:
\(\begin{array}{l}
{C_2}{H_5}OH + {O_2} \to C{H_3}COOH + {H_2}O\\
{V_{{C_2}{H_5}OH}} = \dfrac{{23 \times 750}}{{100}} = 172,5ml\\
\to {m_{{C_2}{H_5}OH}} = D \times V = 138g\\
\to {n_{{C_2}{H_5}OH}} = 3mol\\
\to {n_{C{H_3}COOH}} = {n_{{C_2}{H_5}OH}} = 3mol\\
\to {m_{C{H_3}COOH}} = 180g\\
\to {m_{C{H_3}COOH}}thực tế= \dfrac{{180 \times 80}}{{100}} = 144g\\
C{\% _{C{H_3}COOH}} = \dfrac{{{m_{C{H_3}COOH}}thực tế}}{{{m_{dd}}}} \times 100 = 5\\
\to {m_{dd}} = \dfrac{{{m_{C{H_3}COOH}}thực tế\times 100}}{{C\% }} = 2880g
\end{array}\)