Answer:
A) Mass flow rate of air = 22.892 kmol/hr
B)percentage by mass of oxygen in the product gas = 22.52%
Explanation:
We are given that the mixture containing 9.0 mole% methane in air flowing.
Thus, we have 0.09 mole of methane(CH4) and the remaining will be the air which is (100% - 9%) = 91% = 0.91
Molar mass of CH4 = 12 + 1(4) = 16 g/mol
We are given the average molecular weight of air = 29 g/mol
Thus;
Average molar mass of air and methane mixture is;
M_avg = (0.09 × 16) + (0.91 × 29)
M_avg = 27.83 g/mol
We are told that air flowing at a rate of 7 × 10² kg/h = 700 kg/h
Thus;
Mass flow rate of CH4 in air mixture = 700kg/h × (0.09CH4)/1 mix × (1/27.83kg/kmol) = 2.264 kmol/hr
Mass flow rate of air in mixture = 2.264kmol/h × 0.91kmol air/0.09kmolCH4 = 22.892 kmol/hr
We are told that the mixture is capable of being ignited if the mole percent of methane is between 5% and 15%.
Thus, for 5% of methane, the air required will be;
2.264kmol/h × 0.95kmol air/0.05kmol CH4 = 43.016 kmol/hr
Now, the dilution air needed will be =
43.016 - 22.892 = 20.124 kmol/hr
Total mass flow rate of mixture =
700kg/hr + (20.124kmol/hr × 29kg/mol) = 1283.596 kg/hr
We are told that air consist of 21 mole% Oxygen (O2).
Molar mass of oxygen = 32
Thus;
Mass fraction of oxygen in the product gas = 43.016kmol/h × (0.21molO2/1mol air) × (32kg oxygen/1kmol oxygen) × (1/1283.596kg/h) = 0.2252
Thus, written in percentage form, we have; 22.52%
So, percentage by mass of oxygen in the product gas = 22.52%
= constant
= 
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