Ethene is burnt with 140% of excess air, what is the percentage of O2 in the products?
Theory: we will apply the stochioemtric and mol balance equations to determine our answers
Solution: Step 1:
C2H4 +3O2--> 2CO2 +2H2O
Each mole of C2H4 requires 3 moles and in burning of C2H4 total 4 moles of products are formed
Step 2: excess O2 = 140%
Actual oxygen = required moles of O2*1.4 = 3*1.4 = 4.2
We know that air has 79% mol N2 and 21% O2.
So moles of N2 to system = O2 moles*79/21 = 15.8 moles
Step 3: so for each one mole of C2H4 burnt we have ,
1.2 moles of unreacted O2
15.8 moles of N2 unreacted
2 moles of CO2 formed
2 moles of H2O formed
Total moles of product = 1.2+15.8+2+2=21 moles
O2% in product = 1.2/21*100 = 5.71 mol%
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