Question

Methane is the dominant constituent of the domestic gas supply.

1. Using the lowest excess of air suitable for the fuel, find the amount of air required to burn 1 kmol of methane.

2. Find the mass of air required to burn 1 kg of methane.

3. Find the composition of the exhaust gas. Take LPG as being 92% propane and 8% butane.

4. Using the lowest excess of air suitable for the fuel, find the amount of air required to burn 1 kmol of LPG.

5. Find the mass of air required to burn 1 kg of LPG.

6. Find the composition of the exhaust gas. Acetylene (C2H2) is a gas which, when burnt in pure oxygen, is used in the flame-cutting of metals

7. Using the lowest excess of oxygen suitable for the fuel, find the amount of air required to burn 1 kmol of acetylene.

8. Find the mass of oxygen required to burn 1 kg of acetylene.

9. Why use oxygen instead of air?

Answer 1

Solution:

1) The combustion of methane take place as,

2CH4 + 4O2 = 2CO2 + 4H2O

Since, 2 mol methane required 4 mol of O2.

Hence, 1 kmol of methane requires

= 1 kmol x 4 /2 = 2 kmol of O2

1 mol of O2 = 32 g

Hence, 2 kmol O2 = 2000 mol x 32 g /mol

= 64000 g = 64 kg

2) Since, for burning of 2 mol of CH4, required amount of O2 = 4 mol

Molar mass of CH4 = 16 g

Thus, we can say that

32 g methane requires 4 mol ( 4 x 32 =128 g) of O2

1000 g methane requires = 1000 g x 128/32 = 4000 g of O2

= 4 kg of O2