How many grams of water vapor can be produced if you begin with 86.02 g of C3H8 and 11.20 g of O2?

Answer 1

Looking at the balanced equation for the combustion of C3H8 (propane), we have...

C₃H₈ + 5O₂ ==> 3CO₂ + 4H₂O ... balanced equation

Using dimensional analysis, we can find g of H₂O produced, but first we will determine which, if either, reactant is limiting, i.e. in limiting supply.

Looking at C₃H₈:

86.02 g C₃H₈ x 1 mol C₃H₈ / 44 g x 4 mol H₂O / mol C₃H₈x 18 g /mole H₂O = 140.8 g H₂O

Looking at O₂:

11.20 g O₂ x 1 mol O₂ / 32 g x 4 mol H₂O / 5 mol O₂ x 18 g / mole H₂O = 5.04 g H₂O

So, O₂ is limiting and only 5.04 g of H₂O vapor will be produced.