$$C_aH_bO_cN_d + (4a-b-2c+3d/4) H_2O⇒(4a+b-2c-3d/8) CH_4+ (4a-b+2c+3d/8) CO_2 + dNH_3$$ Solution: Given equation is: $$C_aH_bO_cN_d + (4a-b-2c+3d/4) H_2O ⇒ (4a+b-2c-3d/8) CH_4+ (4a-b+2c+3d/8) CO_2 + dNH_3$$

Given composition is: $C_{55}H_{110}O_{35}N_2$

a=55, b=110, c=35, d=2

Using this coefficient, the relating equation becomes:

$C_{55}H_{110}O_{35}N_2+ 11.5 H_2O ⇒ 31.75CH_4+ 23.25CO_2 + 2NH_3$

$(1358) \quad \quad \quad \quad \quad (207) \quad \quad \quad (508) \quad \quad \quad (1023) \quad \quad (34)$

Mass of methane produced per tonne of waste $= 508*1000/1358 \\ = 374.08 kg/tonne$

Density of methane= 0.7167 kg/m3

Volume of methane gas$= mass/density \\ = 374.08/0.7167 \\ = 521.94 m3/tonne.$