Final answer:
Ammonia (NH3) is the limiting reagent in the reaction with oxygen (O2), and the heat released when 0.2441 mol of NH3 reacts is approximately -69.7 kJ based on the given enthalpy change (ΔH°) of -286 kJ/mol.
Step-by-step explanation:
The student has asked how much heat will be released when 0.2441 mol of NH3 react with 0.20 mol of O2 in the given chemical reaction. The reaction provided is:
4 NH3 (g) + 5 O2 (g) → 4 NO (g) + 6 H2O (g), with the enthalpy change (ΔH°) of -286 kJ/mol given for this reaction. However, the reaction is not balanced with respect to the given reactants.
Since the stoichiometry of the balanced reaction involves 4 moles of NH3 and 5 moles of O2, and 0.2441 mol NH3 is less than the 0.20 mol O2 based on the stoichiometry, NH3 is the limiting reactant. We need to use the amount of NH3 to calculate the heat released. As the reaction shows that 4 moles of NH3 releases -286 kJ of heat, we can set up a proportional relationship to find the heat released by 0.2441 mol of NH3.
ΔH = (-286 kJ/mol NH3) × (0.2441 mol NH3)
After carrying out the multiplication, the heat released (ΔH) for the reaction with 0.2441 mol of NH3 would be approximately -69.7 kJ.