If 20.0 grams of propane are burned with 50.0 grams of oxygen:
What mass of CO₂ is produced?
What mass of water is produced?
What mass of which reactant was excess?
Thank you so much.
See your post above.
To find the mass of CO₂ produced, you need to determine the limiting reactant. The limiting reactant is the one that is completely consumed and determines the amount of product formed.
First, we need to write a balanced chemical equation for the reaction between propane (C₃H₈) and oxygen (O₂):
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
The balanced equation tells us that for every 1 mole of propane, we need 5 moles of oxygen to completely react to form 3 moles of CO₂ and 4 moles of water.
Now, let's calculate the moles of propane and oxygen in the given masses:
Molar mass of propane (C₃H₈) = atomic mass of C(12.01) x 3 + atomic mass of H(1.01) x 8 = 44.1 g/mol
Moles of propane = 20.0 g / 44.1 g/mol = 0.4538 mol
Molar mass of oxygen (O₂) = atomic mass of O(16.00) x 2 = 32.0 g/mol
Moles of oxygen = 50.0 g / 32.0 g/mol = 1.5625 mol
According to the balanced equation, the ratio of propane to oxygen is 1:5. So, for every 0.4538 moles of propane, we need 0.4538 moles x 5 = 2.269 moles of oxygen.
As we have more than enough oxygen, its quantity exceeds the stoichiometric requirement. Thus, propane is the limiting reactant.
To calculate the mass of CO₂ produced, we use the mole ratio between propane and CO₂ from the balanced equation (1:3):
Molar mass of CO₂ = atomic mass of C(12.01) + atomic mass of O(16.00) x 2 = 44.01 g/mol
Moles of CO₂ = 0.4538 mol x 3 = 1.3614 mol
Mass of CO₂ = moles of CO₂ x molar mass of CO₂ = 1.3614 mol x 44.01 g/mol ≈ 59.88 g
Therefore, approximately 59.88 grams of CO₂ are produced.
To find the mass of water produced, we use the same approach as before, considering the molar ratio between propane and water (1:4):
Molar mass of water (H₂O) = atomic mass of H(1.01) x 2 + atomic mass of O(16.00) = 18.02 g/mol
Moles of water = 0.4538 mol x 4 = 1.8152 mol
Mass of water = moles of water x molar mass of water = 1.8152 mol x 18.02 g/mol ≈ 32.72 g
Therefore, approximately 32.72 grams of water are produced.
Finally, to determine the excess reactant, we compare the amount of reactant used to the stoichiometrically required amount. In this case, we compare the moles of oxygen (1.5625 mol) to the amount needed (2.269 moles). Since the moles of oxygen are less than the required amount, propane is the limiting reactant, and oxygen is in excess.
I hope this explanation helps! Let me know if you have any further questions.