Question

How many grams of O2(g) are needed to completely burn 73.7 g of C3H8(g)?

Answer 1

If I remember correctly, you start like this:

If you burn propane, C3H8, you get carbon dioxide and water:

_____ O2 + C3H8 ---> _____ CO2 + _____ H2O , and the number of atoms have to balance out on each side

You have 3 carbon atoms on the left, so you need 3 on the right:

_____O2 + C3H8 ----> 3CO2 + ______ H2O

And we had 8 hydrogen on the left, so we also need 8 hydrogen, or 4H2 on the right:

----O2 + C3H8 ----> 3CO2 + 4H2O

So now we have 10 oxygen on the right, so we need 10 on the left, or 5O2:

5O2 + C3H8 ---> 3CO2 + 4H2O

Then they say that you have 73.7 grams of C3H8. The mass of one mole of C3H8 is 12X3 + 1X8 = 36 + 8 = 44g. So you have 73.7/44 or 1.675 moles of C3H8.

For every mole of C3H8, you need 5 moles of O2. So you will need 5 X 1.675, or 8.375 moles of O2.

Each mole of O2 weighs 16 X 2 or 32 g/mole

So if you need 8.375 moles of O2, that would weigh 8.375 X 32, or 268 grams of O2 ... I think! :) It has been a while since I have had to do any chemistry.