Question

Ethene is burnt with 140% of excess air, what is the percentage of O2 in the...

Ethene is burnt with 140% of excess air, what is the percentage of O2 in the products?

Homework Answers

Answer #1

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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