\(n_{C_2H_5OH}=\dfrac{4,6}{46}=0,1mol\)
\(2C_2H_5OH+2Na\rightarrow2C_2H_5ONa+H_2\)
0,1 0,1 0,05
\(V_{H_2}=0,05\cdot22,4=1,12l\)
\(m_{C_2H_5ONa}=0,1\cdot68=6,8g\)
nC2H5OH=4,6/46=0,1mol
2C2H5OH+2Na→2C2H5ONa+H2
0,1 0,1 0,05
VH2=0,05⋅22,4=1,12l
mC2H5ONa=0,1⋅68=6,8g