This question concerns stoichiometry.

Question

If 17.5 L of methane, CH4, combust completely with excess oxygen at STP, determine the volumes of each of the products form.

Answer 1

No need to go through a lot of calculations if you understand that the stoichiometry of the balanced equations not only refers to moles, but also to volumes (liters) as well as to molecules, etc. So ....

First things first: Write a correctly balanced equation...

CH₄ + 2O₂ ==> CO₂ + 2H₂O

Next, use the coefficients in this balanced equation to find volumes of each of the products:

For CO₂: 17.5 L CH₄ x 1 L CO₂ / L CH₄ = 17.5 L CO₂ formed

For H₂O: 17.5 L CH₄ x 2 L H₂O / L CH₄ = 35.0 L H₂O formed

It's that simple !

Answer 2

In order to answer this question, you must first write out the balanced equation for the combustion of methane in excess oxygen gas:

CH₄ + 2O₂ --> CO₂ + 2H₂o

The problem tells us that the methane combusts completely in excess oxygen- this means we can assume all of the methane is used up to form the products.

Now we are ready for some quick calculations! First, let's figure out how many moles of methane we started with. We know that at STP, 1 mole of any gas occupies 22.4 liters. So, if we divide 17.5 L by 22.4, we know that we started with 0.78125 moles of methane.

We can now use the mole to mole ratios found in the balanced chemical reaction above to find how many moles of each product were produced.

For every mole of methane consumed, one mole of carbon dioxide is produced. This means that 0.78125 moles of carbon dioxide were produced. We can multiple this number by 22.4 Liters/mol to get 17.5 liters of carbon dioxide produced.

For every mole of methane consumed, two moles of water (as a gas) are produced. This means that 1.5625 moles of gaseous water were produced. We can again multiply this number by 22.4 liters/mol to get 35 liters of H₂O produced.