Hi! Not sure where to start with this question. I have provided the answer that
ID: 1819323 • Letter: H
Question
Hi! Not sure where to start with this question. I have provided the answer that I am trying to work towards at the end of this post. Detailed steps would be greatly appreciated.
A certain natural gas has the following volumetric analysis:
This gas is burned with the stoichiometric required amount of dry air The combustion process is found to yield 100% conversion of the hydrogen, but only 85% conversion of the methane. What is the mole percent and weight percent of all constituents (chemical species) in the exhaust gas?
Hint: To get the stoichiometric amount of oxygen required, consider the product gas to contain only CO2, H2O and N2. For the 85% conversion of methane, treat each feed species separatley, meaning that 15% of the methane passes through the reactor and becomes part of the products slate while all of the hydrogen reacts.
ANSWER: CO2 in the product gas represents 8.54 mole% and 13.54 wt% of the total gas.
Explanation / Answer
Basis:100 moles of Feed
1. CH4 + 202 ----------> CO2 + 2H20
55.25 110.5 55.25 110.5
65*0.85= 55.25 moles of CH4 reacts with 2*55.25=110.5 moles of O2 to produce 55.25 moles of CO2 and 110.5 moles of H2O.
2. H2 + 1/2O2 ----------> H2O
8 4 8
Same as above 8 moles reacts with 4 to produce 8 moles.
Total O2 required = 110.5+4-3=111.5 moles (Since we already have 3 moles)
Assuming Air to be a mixture of 79% N2 & 21% O2 (mol %)
Moles of Air required = 111.5/.21=530.95 moles
Moles of N2 from Air = 530.95 * .79 =419.45 moles
Feed:(Moles) Product:(Moles)
CH4-65 CH4-65*.15=9.75
H2-8 N2-18+419.45=437.45
N2-18 CO2-6+55.25=61.25
02-3 H20-8+110.5=118.5
CO2-6 Total=626.95
Product: Mol % Wt(gm) Wt%
CH4 1.55 9.75*16=156 0.90
N2 69.77 437.45*28=12248.6 71.08
CO2 9.77 61.25*44=2695 15.64
H2O 18.90 118.5*18=2133 12.38
Total=17232.6
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