c3h8 + 5o2 --> 4h2o + 3co2

32 grams of propane (c3h8) is burned in excess oxygen gas to produce how many grams of water?

There is 52.33 grams of water produced.

Explanation:

Step 1: Data given

Mass of propane burned = 32.00 grams

Molar mass of propane = 44.1 g/mol

Oxygen is in excess

Molar mass of water = 18.02 g/mol

Step 2: The balanced equation

C3H8 + 5O2 → 4H2O + 3CO2

Step 3: Calculate moles of propane

Moles of propane = mass propane / molar mass of propane

Moles of propane = 32.00 grams / 44.1 g/mol

Moles of propane = 0.726 moles

Step 4: Calculate moles of H2O

Propane is the limiting reactant.

For 1 mol of propane consumed, we need 5 moles of O2 to produce 4 moles of H2O and 3 moles of CO2

For 0.726 moles of propane we'll have 4*0.726 = 2.904 moles of H2O

Step 5: Calculate mass of H2O

Mass of H2O = moles of H2O * molar mass of H2O

Mass of H2O = 2.904 moles * 18.02 g/mol

Mass of H2O = 52.33 grams

There is 52.33 grams of water produced.