Do hiệu suất chỉ đạt $95\%$ :
$ \rightarrow V_{CO_2} = 95\% × 5,6 = 5,32 (l) $
$ n_{CO_2} = \dfrac{V}{22,4} = \dfrac{5,32}{22,4} = \dfrac{19}{80} (mol) $
$ C_6H_{12}O_6 \xrightarrow{\text{Men rượu } , 30-35^oC } 2C_2H_5OH + 2CO_2 $
a,
$ n_{C_2H_5OH} = n_{CO_2} = \dfrac{19}{80} (mol) $
$ \rightarrow m_{C_2H_5OH} = n × M = \dfrac{19}{80} × 46 = 10,925 (g) $
b,
$ n_{C_6H_{12}O_6} = \dfrac{1}{2} × n_{CO_2} = \dfrac{1}{2} × \dfrac{19}{80} = \dfrac{19}{160} (mol) $
$ \rightarrow m_{C_6H_{12}O_6} = n × M = \dfrac{19}{160} × 180 = 21,375 (g) $