How much air is needed to burn 1m3 of gas containing 90% methane, 5% ethane, 3% propane, 2% nitrogen?

Given:

V (gas) = ​​1m3

+ O2

ϕ (CH4) = 90% or 0.9

ϕ (C2H6) = 5% or 0.05

ϕ (C3H8) = 3% or 0.03

ϕ (N2) = 2%

V (air) -?

Decision:

1) Calculate the amount of oxygen consumed for the combustion of methane:

V (CH4) = 1m3 * 0.9 = 0.9m3

0.9 m3 x m3

CH4 + 2 O2 = CO2 + 2 H2O

1 volume 2 volumes

X = 0.9 * 2/1 = 1.8 m3 (O2)

2) Calculate the volume of oxygen consumed for ethane combustion:

V (C2H6) = 1m3 * 0.05 = 0.05 m3

0.05 m3 x m3

2 С2Н6 + 7 О2 = 4 СО2 + 6 Н2О

2 vol. 7 vol.

X = 0.05 * 7/2 = 0.175 m3 (O2)

3) Calculate the volume of oxygen consumed for propane combustion:

V (C3H8) = 1 m3 * 0.03 = 0.3 m3

0.03 m3 x m3

С3Н8 + 5 О2 = 3 СО2 + 4 Н2О

1 vol. 5 vol.

X = 0.03 * 5/1 = 0.15 m3 (O2)

Nitrogen does not burn.

4) Find the volume of oxygen:

V (O2) = 1.8 m3 + 0.175 m3 + 0.15 m3 = 2.125 m3

Oxygen in the air contains 21% or 0.21

V (air) = 2.125 m3 / 0.21 = 10.12 m3

Answer: 10.12 m3 (air).