The molar enthalpy of combustion for ethane, in an open-air environment, is −1428.4kJ/mol. ?
a. Write the balanced thermochemical equation including the term of enthalpy given.
b. What is the molar enthalpy of reaction for CO2 (g) in this equation?
c. What mass of CO2 (g) will be produced if 25.0 MJ of heat is released from this reaction?
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C₂H₆ + (7/2)O₂ → 2CO₂ + 3H₂O ΔH = -1428.4 kJ
When 2 mol of CO₂ is formed, the enthalpy change is -1428.4 kJ.
The molar enthalpy of reaction for CO₂ = (-1428.4 kJ) / (2 mol)= -714.2 kJ/mol
Moles of CO₂ produced = (25.0 × 10⁶ J) / (714.2 × 10³ J/mol) = 35.0 mol
Molar mass of CO₂ = (12.0 + 16.0×2) g/mol = 44.0 g/mol
Mass of CO₂ produced = (35.0 mol) × (44.0 g/mol) = 1540 g