\(2Na+2C_2H_5OH\rightarrow2C_2H_5ONa+H_2\\ n_{C_2H_5OH}=\dfrac{4,6}{46}=0,1\left(mol\right)\\ n_{Na}=n_{C_2H_5OH}=0,1\left(mol\right)\\ n_{H_2}=\dfrac{0,1}{2}=0,05\left(mol\right)\\ a=m_{Na}=0,1.23=2,3\left(g\right)\\ V=V_{H_2\left(đktc\right)}=0,05.22,4=1,12\left(l\right)\)