\(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
PT: \(2Na+2C_2H_5OH\rightarrow2C_2H_5ONa+H_2\)
Theo PT: \(n_{C_2H_5OH}=2n_{H_2}=0,4\left(mol\right)\)
\(\Rightarrow m_{C_2H_5OH}=0,4.46=18,4\left(g\right)\)
\(\Rightarrow V_{C_2H_5OH}=\dfrac{18,4}{0,8}=23\left(ml\right)\)