\(n_{C_2H_4}=\dfrac{33,6}{22,4}=1,5mol\)
\(C_2H_4+H_2O\rightarrow\left(t^o,axit\right)C_2H_5OH\)
1,5 1,5 ( mol )
\(m_{C_2H_5OH}=1,5.46.80\%=55,2g\)
\(n_{C_2H_4}=\dfrac{33,6}{22,4}=1,5\left(mol\right)\)
PTHH: C2H4 + H2O --axit--> C2H5OH
1,5-------------------------->1,5
=> mC2H5OH = 1,5.80%.46 = 55,2 (g)