ASAPPP!!! C3H3(g) + 5 O2 (g) + 3 CO2 (g)+ 4 H20 (g)
What volume of oxygen is required to completely combust 0.632L of propane (C3H8) at STP?

3.16 liters O2

Explanation:

I'll assume the C3H3 is really supposed to read C3H8, for propane.

C3H3(g) + 5 O2 (g) + 3 CO2 (g)+ 4 H20 (g)

C3H8(g) + 5 O2 (g) + 3 CO2 (g)+ 4 H20 (g)

It is unlikely that all the gases involved in this reaction will remain at STP due to the large amount of energy released, we'll assume STP since:

1)  We are told STP and no other information is given, and

2)  STP makes the problem much easier, since all gasses occupy 22.4 liters per mole of that gas.

The balanced equation tells us we need 5 moles of O2 for every 1 mole of C3H3:  a molar ratio of 5/1 (moles O2/moles C3H8).

Calculate moles C3H8 in 0.632 liters of the stuff [metric term for compound]:

  (0.632 liters)/(22.4 liters/mole) = 0.0282 moles C3H8

This means we'll need 5 times that number for the amount of O2 required.  That comes to 0.1411 moles O2.

Convert that to volume:  (0.1411 moles O2)*(22.4 liter/mole) = 3.16 liters O2

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