Final answer:
The limiting reactant is NH₃, which determines the amount of H₂O produced. For 7.50 moles of NH₃, approximately 11.25 moles of H₂O are formed in the reaction according to stoichiometric calculations.
Step-by-step explanation:
To find out how many moles of H₂O are formed when 7.50 mol of NH₃ and 8.20 mol of O₂ react according to the balanced equation 4 NH₃(g) + 5 O₂(g) → 4 NO(g) + 6 H₂O(g), we must perform stoichiometric calculations.
First, we determine the limiting reactant by comparing the mole ratio of the reactants to the coefficients in the balanced chemical equation. According to the equation, it takes 4 moles of NH₃ to react with 5 moles of O₂. Using the provided amounts, the mole ratio for NH₃ to O₂ is 7.50 mol to 8.20 mol, making NH₃ the limiting reactant because it requires 1.375 mol O₂ per mole of NH₃, which is less than the available moles of O₂ (1.64 mol O₂ per mole of NH₃).
Knowing that NH₃ is the limiting reactant, we use its mole ratio with H₂O from the equation to calculate the amount of H₂O produced. From the equation, 4 moles of NH₃ produce 6 moles of H₂O. Therefore, 7.50 moles of NH₃ will produce (7.50 mol NH₃) × (6 mol H₂O)/(4 mol NH₃), which is approximately 11.25 moles of H₂O.