Calculating CO2 Released in Propane BBQ Reaction

[rowkem]

I've changed the numbers a bit so things work out a little easier

Homework Statement

The combustion of propane in a BBQ follows the balanced chemical reaction:

C₃H₈ + 5 O₂ ---> 3 CO₂ + 4 H₂O

As this reaction occurs, approximatley 3000J of heat is released. If a roast takes approx. 1500J of heat to cook and only 10% of the heat is actually used to cook the roast, what mass of CO₂ is released into the atmosphere?

Homework Equations

PV=nRT

The Attempt at a Solution

I'm asking this more as a check as to the process I went through, which was as follows:

1) Figured out how much 10% of 3000J was; 300J
2) Re-wrote the equation in terms of 300J:

0.10C₃H₈ + 0.50 O₂ ---> 0.30 CO₂ + 0.40 H₂O

Using that equation figured out how many moles of CO₂ it would take in order to cook the roast: (1500/300)(0.30) = 1.5 moles

Fgured out how many grams of CO₂ that was: ~66g

Compared that to the mass of CO₂ released in the original equation: ~132

Found the difference between the 2 values: 132-66 = ~66g

So 66g of CO₂ is released into the atmosphere...is that correct?

 

[chemisttree]

I got 15 X 44g or 660 grams of CO2.

Does it actually take 1500J to cook the roast or 10% of 1500J? It wasn't clear from the way you phrased the question but I took it to mean that of the Joules of heat produced in the combustion of propane, only 10% are absorbed by the roast. That means you will need 10 X 1500 J to cook the roast.