Combustion of 5kmol Fuel: Theoretical Air Calculation

[tweety1234]

Homework Statement

A fuel contains 48.0 mol% ethane (C2H6) and the remainder of the fuel is propene (C3H6). Determine the theorectical air in kmol for combustion of 5 kmol of the fuel. Give your answer to an accuracy of one decimal place.

I got 10.2 kmol , is this correct?

the reaction are

C3H8 + 5 O2 ---> 3 CO2 + 4 H2O
2(c2h6) + 7(o2) --> 4(co2) + 6(h2o)

 

[phyzguy]

No, your answer is too small. How many kmol of O2 are needed, and what fraction of the air is O2?

 

[Lame One]

Just making an observation here. The first reaction is unneeded, namely because C3H8 is not the chemical equation of propene. I thought it was a typo at first, but then I noticed that your equation was balanced out. Once you correct that, try again.

 

[tweety1234]

Just making an observation here. The first reaction is unneeded, namely because C3H8 is not the chemical equation of propene. I thought it was a typo at first, but then I noticed that your equation was balanced out. Once you correct that, try again.

Oh right, thanks for pointing that out so the equations are;

$$2C_{2}H_{6} + 7O_{2} = 4CO_{2} + 6H_{2}O$$

$$2C_{3}H_{6} + 9O_{2} = 6CO_{2} + 6H_{2}O$$

we have 5kmol of the fuel, 48% is $$C_{2}H_{6}$$ and 52% is $$C_{3}H_{6}$$

0.48(5) = 2.4 ethene

0.52(5) = 2.6 propene

so to work out the theoretical air, = $$\frac{ Air supplied - theoretical Air}{theoretical air} \times 100$$

from the stiochemtric of the equations, theoretical O2 required is ; (2.6 x 7) + (2.4 x 9) = 39.8 kmol theoretical O2

multiply this by 0.21 , as air contains 21% O2 79% N2,

39.8 X 0.21 = 8.358 kmolNot sure where to go from here, any hints?

 

[phyzguy]

You need to divide by 0.21. Since air is only 21% O2, you need more moles of air to get a certain number of moles of O2. After that, I think you're done.

 

[tweety1234]

You need to divide by 0.21. Since air is only 21% O2, you need more moles of air to get a certain number of moles of O2. After that, I think you're done.

Thank you, But

If I divide by 0.21 I get 189.5? 39.8 / 0.21 = 189.5

The correct answer is 95.7

 

[Jack the Stri]

Remember that you're taking seven (or nine) molecules of oxygen for every two molecules of ethane (propene).

Also, you're mixing up the percentages of ethane and propene.

 

[tweety1234]

Remember that you're taking seven (or nine) molecules of oxygen for every two molecules of ethane (propene).

Also, you're mixing up the percentages of ethane and propene.

Still don't get the right answer, can you please show me how to do this?

Thanks

 

[Jack the Stri]

Your method is correct, just fill in the correct figures and don't forget to divide by two (because for each mole of e.g. ethane you use 3.5 moles of oxygen, not 7!)

 

[tweety1234]

Your method is correct, just fill in the correct figures and don't forget to divide by two (because for each mole of e.g. ethane you use 3.5 moles of oxygen, not 7!)

Oh right, thanks, that's where I was going wrong,

thanks for your help

 

[Jack the Stri]

You're welcome ^^