PT: \(C_2H_6O+3O_2\underrightarrow{t^o}2CO_2+3H_2O\)
Ta có: \(n_{CO_2}=\dfrac{12,395}{24,79}=0,5\left(mol\right)\)
Theo PT: \(n_{C_2H_6O}=\dfrac{1}{2}n_{CO_2}=0,25\left(mol\right)\)
\(\Rightarrow m_{C_2H_6O}=0,25.46=11,5\left(g\right)\)
\(\Rightarrow V_{C_2H_6O}=\dfrac{11,5}{0,8}=14,375\left(ml\right)\)
Độ alcohol = \(\dfrac{14,375}{50}.100=28,75^o\)