Volume of air (moist)

Let X Kg mole moist air

Since the air is moist, we have to calculate composition of air.

P_N2 + P_O2+ P_H2O = 740 mm Hg

P_N2 + P_O2 = 740 – 0.8 × 26

P_N2 + P_O2 = 719.2 mm Hg

P_N2 = 568.168 mm

P_O2 = 151.032 mm

P_H2O = 20.800 mm

Composition of 1 Kg mole of moist air
N₂ = 0.7677
O₂ = 0.2041
H₂O = 0.0281

N₂ balance

N in coal + N₂ from moist air = N₂ in Producer gas
0.017/28 + 0.7677X = 0.53 × 0.208
X = 0.14279 Kg mole  or = 3.601 m³ (26^oC and 740 mm Hg)

Weight of steam : Hydrogen balance

Consider Z Kg mole steam.
0.025 + 0.00094 + Z + 0.00401 = 0.004472
Z = 0.015 Kg mole
   = 0.266 Kg steam/Kg coal

% H₂O blown in, that was decomposed

Water vapour in PG = Water from evaporation of M of coal + Water of undecomposed steam
   0.025 × 0.208 = 0.017/18 + W
   W = 0.004255 Kg mole undecomposed steam
Steam decomposed = {0.266 − (0.004255 × 18)}
                              = 0.1895 Kg

% steam blown, that is decomposed in producer gas = 0.1895 × 100/ 0.266
                                                                             = 71.2