Madi H.

asked • 09/13/17

Help with Chemistry problem

How many grams of carbon dioxide can form when a mixture of 4.41 g propane (C3H8) and 6.40 g of oxygen (O2) is ignited, assuming complete combustion to form carbon dioxide and water.

The balanced equation is:

C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l)
 
I solved the problem and ended up with an answer of 13.2 grams. If you could please check the answer for me and see if I am solving it correctly, I would definitely appreciate it.

1 Expert Answer

By:

Shane E. answered • 09/13/17

Tutor
4.9 (169)

4+ years experience tutoring and 2 years as a General Chemistry TA

Madi H.

Thank you! I followed all of the steps you listed and I saw where I was messing up at. I was using the wrong limiting reactant, so this time I got 10.6 grams as the answer. 
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09/13/17

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 C3Hx 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

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