a, \(n_{C_2H_4}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
PT: \(C_2H_4+H_2O\underrightarrow{^{t^o,xt}}C_2H_5OH\)
Theo PT: \(n_{C_2H_5OH\left(LT\right)}=n_{C_2H_4}=0,15\left(mol\right)\)
Mà: H = 80%
\(\Rightarrow n_{C_2H_5OH\left(TT\right)}=0,15.80\%=0,12\left(mol\right)\)
\(\Rightarrow m_{C_2H_5OH\left(TT\right)}=0,12.46=5,52\left(g\right)\)
b, \(V_{C_2H_5OH}=\dfrac{5,52}{0,8}=6,9\left(ml\right)\)