Question

A mixture of methane and air is capable of being ignited only if the mole percent of methane is between 5% and 15%. A mixture containing 9.00 mole% methane in air flowing at a rate of 700.0 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 21.0 mole% O2 and 79.0 mole% N2 and hence to have an average molecular weight of 29.0).

Answer 1

The required flow rate of air in mol/h is 20144 and the percent by mass of oxygen in the product gas is 23,1%.

Step-by-step explanation:

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

`0,09*16,04g/mol + 0,91*29g/mol = 27,8g/mol`

And a flow of 700'000g/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.

Solving, X = 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 mass of oxygen in the product gas is: =0,231 kg O₂/ kg air* 100 = 23,1%

I hope it helps!