a, Ta có: \(V_{C_2H_5OH}=150.96\%=144\left(ml\right)\)
\(\Rightarrow m_{C_2H_5OH}=144.0,8=115,2\left(g\right)\)
\(\Rightarrow n_{C_2H_5OH}=\dfrac{115,2}{46}\approx2,5\left(mol\right)\)
PT: \(2C_2H_5OH+2Na\rightarrow2C_2H_5ONa+H_2\)
Theo PT: \(n_{H_2}=\dfrac{1}{2}n_{C_2H_5OH}=1,25\left(mol\right)\Rightarrow V_{H_2}=1,25.22,4=28\left(l\right)\)
b, Độ rượu = \(\dfrac{144}{150+50}.100=72\) (độ)