the combustion of 0.1240 kg of propane in the presence of excess oxygen produces 0.3110kg of carbon dioxide what is the limittinf reacrant?

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%.

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.