Ta có:
\(V_{C2H5OH}=0,2.96\%=0,192\left(l\right)=192\)
\(\Rightarrow m_{C2H5OH}=192.0,8=153,6\left(g\right)\)
\(\Rightarrow n_{C2H5OH}=\frac{153,6}{46}=3,339\left(mol\right)\)
\(n_{C2H5OH\left(lt\right)}=\frac{3,339}{90\%}=3,71\left(mol\right)=V_{C2H4}=3,21.22,4=83,104\left(l\right)\)