C3h8 + 5o2 = 3co2 + 4h2o when 44.0 grams of propane (c3h8) under goes complete combustion, how many grams of water will be produced?

72 grams of water will be produced.

Explanation:

To calculate the number of moles, we use the formula:

   ....(1)

Mass of propane = 44 grams

Molar mass of propane = 44 grams

Putting values in above equation, we get:

For the reaction of combustion reaction of propane, the equation follows:

By Stoichiometry of the reaction,

1 mole of propane produces 4 moles of water.

So, 1 mole of propane will produce = of water.

Now, to calculate the amount of water, we use equation 1, we get:

Molar mass of water = 18 g/mol

Mass of water produced = 72 grams

Hence,  72 grams of water will be produced.