$$C_aH_bO_cN_d + (4a+b-2c-3d/4) O_2 ⇒ aCO2 + (b-3d/2) H2O +dNH3$$

$Solution: \\ Step \ \ 1: Determine \ \ the \ \ oxygen \ \ requirements: \\ Given \ \ equation \ \ is:$

$$C_aH_bO_cN_d + (4a+b-2c-3d/4) O_2 ⇒ aCO2 + (b-3d/2) H2O +dNH3$$

Given composition: $C_{55}H_{80}O_{30}N_2$

The required coefficients are: a=55, b=80, c=30, d=2

Using the coefficient we get the equation:

$C_{55}H_{80}O_{30}N_2 + 58.5 O_2 ⇒ 55CO_2 + 37H_2O +2NH_3 \\ (1248) \hspace{1.5cm}(1872) \hspace{0.8cm}(2420) \hspace{0.7cm}(666) \hspace{0.6cm}(34)$

Oxygen required /tonne$= 1872*1000/1248 \\ = 1500 kg/tonne…………………..1$

Step 2: Determine the oxygen required to stabilize ammonia:

$NH_3 + 2O_2 ⇒ H_2O +HNO_3 \\ (17)\hspace{0.6cm}(64)\hspace{0.5cm}(18)\hspace{0.5cm}(63)$

Amount of oxygen required$= 64*1000/1248 \\ = 51.28 kg/tonne…………………….2$

Step 3: Determination of amount of air requirement:

Assume 23.15% oxygen by weight

Density of air= 1.2928 kg/m3

Total amount of oxygen$= 1500+ 51.28 \\ =1551.28 kg/tonne$

Mass of air required$= 1551.28/0.2315 \\ = 6701 kg/tonne$

Volume of air required$= 6701/1.2928 \\ = 5183.32 m3/tonne$