Answer to Question #164225 in Physical Chemistry for Rujul

Question

A gaseous fuel has the following composition by volume c2h4=20%,c4h10=30%,co=15%,o2=2%.calculate the volume of air supplied per m 3 of fuel

Answer 1

Suppose the conditions are normal: T = 20 ^OC = 293 ^OK, P = 101.325 kPa. Also, the air contain 21% O₂. Also, 20%+30%+15+2% = 67%. Let's say the rest 33% is an inflamable gas, for example N₂.

n_gas = $\frac{PV}{RT}$ = $\frac{101.325 \times 1000}{8.314 \times 293}$ = 40.90 (mol)

n(C₂H₄) = 40.90 $\times$ 0.2 = 8.18 (mol)

n(C₄H₁₀) = 40.90 $\times$ 0.3 = 12.27 (mol)

n(CO) = 40.90 $\times$ 0.15 = 6.13 (mol)

n(O₂) = 40.90$\times$0.02 = 0.82 (mol)

C₂H₄ + 3O₂ = 2CO₂ + 2H₂O

C₄H₁₀ + 6$\frac{1}{2}$O₂ = 4CO₂ + 5H₂O

CO + $\frac{1}{2}$O₂ = CO₂

n_O2(needed) = (8.18 $\times$ 3 + 12.27 $\times$ 6.5 + 6.13 $\times$ 0.5) - 0.82 = 106.54 (mol)

V_O2(needed) = 106.54$\times$ 22.4 = 2386.5 (L)

V_air(needed) = 2386.5 / 0.21 = 11364.3 (L) = 11.36 m³

Answer: V_air(needed) = 11.36 m³.