Lên men 90 gam glucozơ, thu được V ml ancol etylic (D = 0,8 gam/ml) với hiệu

PTPƯ: ${C_6}{H_{12}}{O_6}$ $\xrightarrow{Lên men}$ $2{C_2}{H_5}OH$ + $2C{O_2}$
${n_{{C_2}{H_5}OH}} = \,\frac{{90}}{{100}}.2{n_{{C_6}{H_{12}}{O_6}}} = \,0,9\,mol$
→ ${m_{{C_2}{H_5}OH}} = \,41,4g\, \to V\, = \,\frac{{{m_{{C_2}{H_5}OH}}}}{D} = \,51,75ml$