Answer:
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: [tex]\frac{289kg}{1249 kg}[/tex] =0,231 kg O₂/ kg air
I hope it helps!
Answer:
A) Mass flow rate of air = [tex]22.982[/tex] kmol/hr
B) percentage by mass of oxygen in the product gas = [tex]$22.52 \%$[/tex]
Explanation:
We are given that the mixture containing 9.0 mole% methane in air flowing.
Thus, we have [tex]0.09[/tex] mole of methane(CH4) and the remaining will be the air which is [tex]$(100 \%-9 \%)=91 \%$=0.91[/tex]
We are given the average molecular weight of air = [tex]29[/tex] g/mol Thus; Average molar mass of air and methane mixture is;
M-air- [tex]$=(0.09 \times 16)+(0.91 \times 29)$[/tex]
[tex]=$27.83 \mathrm{~g} / \mathrm{mol}$[/tex]
Mass fraction of oxygen in the product gas= [tex]=43.016 \mathrm{kmol} / \mathrm{h} \times(0.21 \mathrm{molO} 2 / 1 \mathrm{~mol}$ air $) \times(32 \mathrm{~kg}$ oxygen/1kmol oxygen $) \times(1 / 1283.596 \mathrm{~kg} / \mathrm{h})=0.2252$[/tex]
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