Ethane burns in oxygen to form CO₂ and H₂O according to the equation: 2C₂H₆ + 7O₂ ⟶ 4CO₂ + 6H₂O If 1250 cc of oxygen is burnt with 300 cc of ethane. Calculate:
the volume of CO₂ formed.
Topic: Organic chemistry: structure, nomenclature and reactions
Answer
2C₂H₆ + 7O₂ ⟶ 4CO₂ + 6H₂O
All the substances of interest are gases at the same temperature and pressure, so the coefficients give the volume ratio directly.
First check which reactant runs out. C₂H₆ : O₂ = 2 : 7 300 cc of C₂H₆ needs (7 ÷ 2) × 300 = 1050 cc of O₂ 1250 cc is available, so oxygen is in excess and ETHANE is the limiting reactant.
Volume of CO₂: C₂H₆ : CO₂ = 2 : 4 = 1 : 2 volume of CO₂ = 2 × 300 = 600 cc
Volume of CO₂ formed = 600 cc
Whenever a question gives you amounts of TWO reactants, one of them will run out first and it alone decides how much product forms. Find it before you do anything else: work out how much oxygen the ethane needs, and compare with how much there is.
Here 300 cc of ethane needs 1050 cc of oxygen and 1250 cc is available, so the ethane is used up while oxygen is left over. All the product calculations run from the 300 cc of ethane. Running them from the 1250 cc of oxygen instead — 1250 × 4 ÷ 7 = 714 cc — is the standard wrong answer.