28.7 l of propane, c3h8, are consumed in a combustion reaction. how many grams of water are produced?

The chemical reaction would be:

C3H8 + 5O2 = 3CO2 + 4H2O

For this case, we assume that gas is ideal thus in every 1 mol the volume would be 22.41 L. We calculate as follows:

28.7 L C3H8 (1 mol / 22.41 L) (4 mol H2O / 1 mol C3H8) (18.02 g / mol) = 92.31 g H2O produced

Hope this answers the question.