ta có : n(etanol) = 0,1(mol)
PTHH :
\(2C2H5OH+2Na->2C2H5ONa+H2\uparrow\)
0,1mol...........................................................0,05mol
a) VH2(đktc) = 0,05.22,4 = 1,12(l)
b) Ta có : V(rượu) = 4,6/0,8 = 5,75(ml)
c) ta có
V(dd rượu) = 5,75 + 8,25 = 14(ml)
=> độ rượu = \(\dfrac{5,75}{14}.100=41^0\)
\(PTHH:2C_2H_5OH+2Na\rightarrow2C_2H_5ONa+H_2\)
\(n_{Na}=\dfrac{m}{M}=\dfrac{4,6}{23}=0,2\left(mol\right)\)
\(\Rightarrow n_{H_2}=0,1\left(mol\right)\)
a) \(V_{H_2}=n.22,4=0,1.22.4=2,24\left(l\right)\)