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The combustion of 0.1240 kg of propane in the presence of excess oxygen produces 0.3110 kg of carbon dioxide. What is the limiting reactant?

PLEASE I REALLY NEED HELP!!!!

User Hugoware
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Answer:

The limiting reactant is the propane gas, C₃H₈ while the percentage yield is 83.77%

Step-by-step explanation:

Here we have

Propane gas with molecular formula C₃H₈, molar mass = 44.1 g/mol combining with O₂ as follows

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

Therefore, 1 mole of C₃H₈ combines with 5 moles of O₂ to produce 3 moles CO₂ and 4 moles of H₂O

Mass of propane = 0.1240 kg = 124.0 g

Number of moles of propane = mass of propane/(molar mass of propane)

The number of moles of propane = 124/44.1 = 2.812 moles

The molar mass of CO₂ = 44.01 g/mol

Mass of CO₂ = 0.3110 kg = 311.0 g

Therefore, number of moles of CO₂ = mass of CO₂/(molar mass of CO₂)

The number of moles of CO₂ = 311.0 kg/ 44.01 g/mol = 7.067 moles

Therefore, since 1 mole of propane produces 3 moles of CO₂, 2.812 moles of propane will produce 3 × 2.812 moles or 8.44 moles of CO₂

Therefore;

The limiting reactant is the propane gas, C₃H₈, since the oxygen is in excess

Hence


The \ percentage \ yield = (Actual \, yield)/(Theoretical \, yield) * 100 = (7.067)/(8.44) * 100 = 83.77 \%

The percentage yield = 83.77%.

User SeanCocteau
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