PTHH: \(2C_2H_5OH+2Na\rightarrow2C_2H_5ONa+H_2\)
Ta có: \(V_{C_2H_5OH}=200\cdot45\%=90\left(ml\right)=0,09\left(l\right)\)
\(\Rightarrow n_{C_2H_5OH}=\frac{0,09}{22,4}=\frac{9}{2240}\left(mol\right)\)
\(\Rightarrow n_{H_2}=\frac{9}{4480}mol\) \(\Rightarrow V_{H_2}=\frac{9}{4480}\cdot22,4=0,045\left(l\right)\)