Amixture of methane and air is capable of being ignited only if the mole percent of methane is between 5% and 15%. a mixture containing 9.0 mole% methane in air flowing at a rate of 700 kg/h is to be diluted with pure air to reduce the methane concentration to the lower flammability limit. calculate the required flow rate of air in mol/h and the percent by mass of oxygen in the product gas. (note: air may be taken to consist of 21 mole% o and 79% n2 and to have an average molecular weight of 29.0.)
(a)n = 20,200 mol air/h; 0.225 kg o2/kg

Explanation:

The air 9% mole% methane have an average molecular weight of:

9%×16,04g/mol + 91%×29g/mol = 27,8g/mol

And a flow of 700000g/h÷27,8g/mol = 25180 mol/h

In the reactor where methane solution and air are mixed:

In = Out

Air balance:

91% air×25180 mol/h + 100% air×X = 95%air×(X+25180)

Where X is the flow rate of air in mol/h = 20144 mol air/h

The air in the product gas is

95%×(20144 + 25180) mol/h = 43058 mol air× 21%O₂ = 9042 mol O₂ ×32g/mol = 289 kg O₂

43058 mol air×29g/mol 1249 kg air

Percent of oxygen is: =0,231 kg O₂/ kg air

I hope it helps!

the answer is d. 74.02