\(2C_2H_5OH+2K\rightarrow2C_2H_5OK+H_2\)
2 1 2 1 (mol)
0,0101 --->0,0101->0,0101--->0,00505 (mol)
V rượu = 138 ml = 0,138 (l)
CT : V = m/D
=> m = V.D
=> m rượu = 0,138 . 0,8 = 0,1104 (g)
m nước = 0,138 . 1 = 0,138 (g)
n rượu = 0,1104 / 46 = 0,0024 (mol)
n nước = 0,138 / 18 = 0,0077 (mol)
nC2H5OH = 0,0024 + 0,0077 = 0,0101 ( mol)
\(mC_2H_5OK=0,0101.84=0,8484\left(g\right)\)
\(VH_2=0,00505.22,4=0,11312\left(l\right)\)