
Shane E. answered 09/13/17
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Hi Madi,
So instead of giving you the answer, I want to see if I can walk you through the problem. This is a limiting reagent/theoretical yield problem. So the first step is to convert both masses (4.41g propane and 6.40g oxygen) to their respective amount in moles. Then you can use the mole ratios (since you have correctly balanced the equation already) to determine the limiting reactant. You do this by converting the moles of propane (calculated from 4.41g divided by the molecular weight of propane) to moles of carbon dioxide (since you want to know how much of CO2 can be formed - if the question asked how much water could be formed, you would convert to moles of H2O using the mole ratios).
Whichever reactant (propane or oxygen) forms the lesser amount of moles of CO2 is your limiting reactant and therefore, it will run out before the other reactant (the amounts given are not perfect ratios and either propane or oxygen will run out before the other). Once you have determined the limiting reactant, you use the number of moles of CO2 that the limiting reactant can form and convert this amount to grams of CO2 using the molecular weight of carbon dioxide. This mass in grams will be the answer and what is called the theoretical yield. The theoretical yield is how much product you can form based the limiting reactant, which you determine based on the given amounts of the reactants.
Does this make sense? If not, please respond and I will try to explain it a different way. Also please respond with your new answer so that I can confirm that you understand how to complete this problem and others like it.
Hope this helped!

Shane E.
I am getting 5.28g, essentially half of what you are getting, so my guess is that you are forgetting a factor of 2 somewhere. Below I have listed how I would perform the problem:
4.41g C3H8 x 1 mol C3H8/44.09 g/mol C3H8 =0.100 mol of C3H8
0.100 mol C3H8 x 3 mol CO2/1 mol C3H8= 0.300 mol CO2
6.40g O2 x 1 mol O2/32 g/mol O2 = 0.200 mol of O2
0.200 mol O2 x 3 mol CO2/5 mol O2 = 0.12 mol CO2
Limiting Reactant = O2
0.12 mol CO2 (formed from oxygen) x 44.01 g CO2/1 mol CO2 = 5.28g CO2
See if you can find your mistake - I think you may have used the wrong molecular weight for Oxygen (32 g/mol, not 16 g/mol)
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09/13/17
Madi H.
09/13/17