Material balance with a reaction (chemical engineering)

[MickeyBlue]

Homework Statement

Felder & Rousseau 4.73 (p. 184)

A mixture of propane and butane is burned with pure oxygen. The combustion products contain 47.4 mole % H2O. After all of the water is removed from the products, the residual gas contains 69.4 mole% CO2 and the balance is O2. What is the mole percent of propane in the fuel?

Homework Equations

1. Let propane = P
2. Choose 100 mol of the reactant fuel as a basis

The Attempt at a Solution

[/B]
Set up 2 balanced equations: 1. C3H8 + 5O2 -> 4H2O + 3CO2
2. C4H8 + 13/2O2 -> 5H2O + 4CO2

I assumed a 100% conversion of the reactant gases. I know O2 is in excess. I drew up a reactant table in terms of mol and set Propane in = P and Butane in= 100-P. I then tried to calculate P by setting the mols of water out divided by the total mols out equal to 0.474.

 

[Chestermiller]

So, what's the problem?

 

[collinsmark]

Hello @MickeyBlue! Welcome to Physics Forums!

Homework Statement

Felder & Rousseau 4.73 (p. 184)

A mixture of propane and butane is burned with pure oxygen. The combustion products contain 47.4 mole % H2O. After all of the water is removed from the products, the residual gas contains 69.4 mole% CO2 and the balance is O2. What is the mole percent of propane in the fuel?

Homework Equations

1. Let propane = P
2. Choose 100 mol of the reactant fuel as a basis

The Attempt at a Solution

[/B]
Set up 2 balanced equations: 1. C3H8 + 5O2 -> 4H2O + 3CO2
2. C4H8 + 13/2O2 -> 5H2O + 4CO2

I assumed a 100% conversion of the reactant gases. I know O2 is in excess. I drew up a reactant table in terms of mol and set Propane in = P and Butane in= 100-P. I then tried to calculate P by setting the mols of water out divided by the total mols out equal to 0.474.

So far so good, with one exception:

You wrote that the butane was $\mathrm{C_4 H_8}$, but I think you meant $\mathrm{C_4 H_{10}}$. Your equations were balanced correctly though, so I assume that was just a minor mistake typing things into the computer.

As a hint for moving forward, don't forget to include the leftover $\mathrm{O_2}$ in your product. Similar to what you did with $P$, make up a variable name to indicate the amount of $\mathrm{O_2}$ in there. I used the variable $x$, but you can use whatever you want.

So your formula, as you have already stated it, should be of this form:

$$\frac {\mathrm{moles \ of \ H_2O}}{(\mathrm{moles \ of \ H_2O}) + (\mathrm{moles \ of \ CO_2}) + (\mathrm{moles \ of \ O_2})} = 0.474$$

After that, take out the $\mathrm{H_2O}$ and do something quite similar except with $\mathrm{CO_2}$ in the numerator (and the ratio value being different).

If you do things right, you'll have two equations and two unknowns ($P$ and $x$). That's enough to solve for $P$.

 

[MickeyBlue]

So, what's the problem?

My final answer for propane was well above 100 and I couldn't understand why.

 

[collinsmark]

My final answer for propane was well above 100 and I couldn't understand why.

Show your work, and perhaps we may be able to point you in the right direction.

 

[MickeyBlue]

Oh! Thank you, I see where I went wrong now. Because of how I added mols of O₂ consumed I took it as the amount in excess instead. You're right; this should be enough to get P.