What is the mass of the water produced if 5.00 grams of propane are consumed in the following reaction?

Question

What is the mass of the water produced if 5.00 grams of propane are consumed in the following

reaction?

C3H8+5O2=3CO2+4H2O

Answer 1

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

-- Convert 5.00 g propane (C₃H₈) to moles of propane using the molar mass of propane of 44.1 g/mol:

5.00 g C₃H₈ x 1 mol / 44.1 g = 0.1134 moles C₃H₈

-- Use moles of propane to calculate moles of H₂O produced by using the mole ratios in the balanced eq.:

0.1134 mols C₃H₈ x 4 mols H₂O / 1 mol C₃H₈ = 0.4536 moles H₂O produced

-- Finally, convert moles of H₂O to grams of H₂O using the molar mass of H₂O of 18 g/mol:

0.4536 mols H₂O x 18 g H₂O / mol = 8.165 moles H₂O = 8.17 g H₂O (3 significant figures)

Answer 2

Three step process for stoichiometry g of i to moles of i to moles of j to grams of j where i = propane and j = H₂O

Start with what you know and convert

5 g P * (mole/g P) (4 moles H₂O/1 mole P) (g/mole H₂O) where g/mole are molar masses and moles/g is 1/molar mass.