Đáp án: $71,875ml$

Giải thích các bước giải:

$n_{CO_2}=\dfrac{33,6}{22,4}=1,5(mol)$

Bảo toàn $C$:

$n_{CO_2}=6n_{C_6H_{12}O_6}$

$\to n_{C_6H_{12}O_6}=0,25(mol)$

$C_6H_{12}O_6\to 2CO_2+2C_2H_5OH$

$\to n_{C_2H_5OH}=0,25.2=0,5(mol)$

$\to m_{C_2H_5OH}=46.0,5=23g$

$\to V_{C_2H_5OH}=23.0,8=28,75ml$

$\to V_{\text{rượu $40^o$}}=28,75:40\%=71,875ml$

Đáp án:

\({V_{dd{\text{ rượu}}}} =71,875{\text{ ml}}\)

Giải thích các bước giải:

 Phản ứng xảy ra:

\({C_6}{H_{12}}{O_6} + 6{O_2}\xrightarrow{{{t^o}}}6C{O_2} + 6{H_2}O\)

\({C_6}{H_{12}}{O_6}\xrightarrow{{men}}2{C_2}{H_5}OH + 2C{O_2}\)

Ta có:

\({n_{C{O_2}}} = \frac{{33,6}}{{22,4}} = 1,5{\text{ mol}}\)

\( \to {n_{{C_6}{H_{12}}{O_6}}} = \frac{1}{6}{n_{C{O_2}}} = 0,25{\text{ mol}}\)

\( \to {n_{{C_2}{H_5}OH}} = 2{n_{{C_6}{H_{12}}{O_6}}} = 0,5{\text{ mol}}\)

\( \to {m_{{C_2}{H_5}OH}} = 0,5.46 = 23{\text{ gam}}\)

\( \to {n_{{C_2}{H_5}OH}} = \frac{{23}}{{0,8}} = 28,75{\text{ ml}}\)

\( \to {V_{dd{\text{ rượu}}}} = \frac{{28,75}}{{40\% }} = 71,875{\text{ ml}}\)