Final answer:
To calculate the heat released, we must use the balanced chemical equation and apply the stoichiometry to the provided 0.1751 mol of NH₃ and ΔH° of -286 kJ/mol, assuming NH₃ is the limiting reactant. The provided reaction equation is unbalanced, so a correct calculation requires the balanced equation, which is missing.
Step-by-step explanation:
To calculate the quantity of heat released when 0.1751 mol of NH₃ are mixed with 0.200 mol of O₂, we need to look at the stoichiometry of the balanced chemical reaction and the enthalpy change (ΔH°) provided.
The given reaction is:
4 NH₃ (g) + 5 O₂ (g) → 4 NO (g) + 6 H₂O (g)
The ΔH° for the formation of 2 moles of N₂H₄ (hydrazine) and 2 moles of H₂O (g) is -286 kJ.
However, the reaction provided is not balanced as written, assuming the products are NO and H₂O. To use ΔH°, we need a balanced chemical equation, which appears to be missing from the question. If we correct and balance the equation, we can then use the balanced equation to calculate the heat released per mole, taking into consideration the given ΔH° and the amount of limiting reactant.
Assuming that NH₃ is the limiting reactant and given that we have 0.1751 mol of NH₃, we can calculate the heat released using the stoichiometry of the correctly balanced equation:
ΔH (reaction) = (Moles of limiting reactant / Moles needed as per balanced equation) × ΔH°
Without the correct balanced equation, we cannot provide the exact quantity of heat released.