Propane, C3H8, reacts with oxygen according to the following balance equation. If 44 grams of propane react completely with sufficient oxygen, how many grams of water are produced?
To find the number of grams of water produced when 44 grams of propane react completely, we need to use the stoichiometry of the balanced chemical equation. The balanced equation for the reaction of propane (C3H8) with oxygen (O2) to produce carbon dioxide (CO2) and water (H2O) is:
C3H8 + 5O2 -> 3CO2 + 4H2O
From the equation, we can see that one mole of propane reacts with 4 moles of water. To calculate the amount of water produced, we need to convert grams of propane to moles using its molar mass and then use the mole ratio to determine the moles of water produced. Finally, we will convert the moles of water back to grams.
1. Determine the molar mass of propane (C3H8):
C = 12.01 g/mol * 3 = 36.03 g/mol
H = 1.01 g/mol * 8 = 8.08 g/mol
Total molar mass of propane: 36.03 g/mol + 8.08 g/mol = 44.11 g/mol
2. Calculate the moles of propane:
moles = mass / molar mass
moles = 44 g / 44.11 g/mol ≈ 1 mole
3. Use the mole ratio from the balanced equation:
1 mole of propane produces 4 moles of water.
4. Calculate the moles of water produced:
moles of water = moles of propane * mole ratio
moles of water = 1 mol * 4 mol H2O / 1 mol C3H8 = 4 moles
5. Convert the moles of water to grams:
mass = moles of water * molar mass of water
mass = 4 moles * (2.02 g/mol) ≈ 8.08 grams
Therefore, when 44 grams of propane react completely with sufficient oxygen, approximately 8.08 grams of water are produced.
44 g is one mole of propane
so four moles of water are produced
the molar mass of water is 18 g