`n_{\text{rượu}}=\frac{18,4}{46}=0,4(mol)`
`2C_2H_5OH+2Na\to 2C_2H_5ONa+H_2`
`a)` `n_{Na}=n_{\text{rượu}}=0,4(mol)`
`=> m_{Na}=0,4.23=9,2g`
`b)` `n_{H_2}=0,5n_{Na}=0,2(mol)`
`=> V_{H_2}=0,2.22,4=4,48(l)`
`c)` Độ rượu = `\frac{V_{\text{ancol etylic}}}{V_{\text{ancol etylic}}+V_{H_2O}}.100`
`=> ` Độ rượu= $\dfrac{\dfrac{18,4}{0,8}}{115}.100$
`=>` Độ rượu=`20^0`