Answer:
Propane
Explanation:
From the question given, we were told that 0.1240 kg of propane reacted with excess oxygen to produce 0.3110kg of carbon dioxide.
Since the reaction took place in the presence of excess oxygen, therefore, propane is the limiting reactant as all of it is used up in the presence of excess oxygen.
Answer:
The limiting reactant is propane, C₃H₈ and the percentage yield is 83.77%
Explanation:
Mass of propane = 0.1240 kg = 124 g
Mass of carbon dioxide = 0.3110 kg = 311 g
Molar mass of propane = 44.1 g/mol
Molar mass of carbon dioxide = 44.01 g/mol
Number of moles of propane = 124/44.1 = 2.812 moles
Number of moles of carbon dioxide = 311/44.01 = 7.067 moles
Equation for the reaction
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
Hence 1 mole of propane ideally yields 3 moles of CO₂
Hence, 2.812 moles of propane will yield 3×2.812 moles = 8.44 moles of CO₂
Since, oxygen is in excess, therefore, the limiting reactant = Propane, C₃H₈
The percentage yield = 7.067/8.44× 100 = 83.77%.
