2C8H18 + 25O2 -> 16CO2 + 18H20 Answer the following question: if 10 moles of water were produced in this reaction, what would be the molar volume?

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The first thing to do when looking at a reaction is to make sure it is balanced, which this one is.  Once we know it is balanced, we can look at the coefficients to determine the molar ratios between the substances.  From our balanced reaction, we see that 2 moles of octane reacts with 25 moles of oxygen to produce 16 moles of carbon dioxide and 18 moles of H2O. 

Since we are given the moles of H2O produced, we can use the stoichiometric ratios to determine the moles of CO2 produced as both the water and the CO2 will be in the gas phase.  From the information given, we must assume that we have no excess oxygen gas left over.

10 mol H2O (16 mol CO2 / 18 mol H2O) = 8.9 moles of CO2 produced

Therefore we have 18.9 moles of gaseous products.  To find the volume of this, we would need to know the temperature and pressure.  If we assume STP (standard temperature and pressure), then the molar volume will be 22.4 L/mol of gas.  Therefore the total volume can be calculated by

18.9 moles ( 22.4 L /mol) = 423 L

If we are at a different temperature and pressure, then the molar volume will be different and therefore the volume of the products will be different.

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