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rowkem
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I've changed the numbers a bit so things work out a little easier
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?
PV=nRT
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?
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?