The complete combustion of propane, C₃H₈(g) , is represented by the equation: C₃H₈(g) + 5 O₂(g) → 3 CO₂(g) + 4 H₂O(l) ΔH° = -2220 kJ
How much heat is evolved in the complete combustion of 20.0 L C₃H₈(g) at STP?

Question

Quizplus · Accepted Answer

Step 1: Calculate the moles of propane at STP. Using the ideal gas law: PV = nRT where P = 1 atm, V = 20.0 L, n = ?, R = 0.08206 L atm/mol K, T = 273 K n = PV/RT = (1 atm)(20.0 L)/(0.08206 L atm/mol K)(273 K) = 0.962 mol C₃H₈ Step 2: Use stoichiometry to find the moles of H₂O(l) produced. From the balanced equation, 1 mol C₃H₈ reacts with 5 mol O₂ to produce 4 mol H₂O(l) and 3 mol CO₂. So, 0.962 mol C₃H₈ will react with 4/3 x 0.962 x 5 = 6.43 mol O₂ Using the volume of 1 mol of any gas at STP (22.4 L), the volume of 6.43 mol of O₂ is: V = n x V_m = 6.43 mol x 22.4 L/mol = 144 L So, the reaction will produce 4/3 x 0.962 x 22.4 x 5 = 61.5 L of H₂O(l) Step 3: Calculate the heat evolved using the given ΔH° value. The heat evolved in the reaction is -2220 kJ per 1 mol of C₃H₈. Therefore, the heat evolved in the reaction for 0.962 mol of C₃H₈ is: -2220 kJ/mol x 0.962 mol = -2135 kJ (Note the negative sign indicates that the reaction is exothermic.) Step 4: Convert the heat evolved to the amount per litre of propane at STP. The heat evolved per litre of propane at STP is: -2135 kJ/20.0 L = -107 kJ/L (Note that the negative sign means that the reaction releases heat.) Therefore, the answer is D: 1982 kJ.

The Complete Combustion of Propane, C3H8(g), Is Represented by the Equation

The Complete Combustion of Propane, C₃H₈(g), Is Represented by the Equation