Final Answer:
The balanced reaction equation indicates that 2 moles of MnO₂ react with 4 moles of KOH and produce 2 moles of H₂O. By using the given masses of MnO₂ (166.0 g), KOH (222.1 g), and O₂ (28.15 g), the limiting reactant will be KOH. With 222.1 g of KOH, 2 moles of H₂O will be produced. Hence, the mass of H₂O formed will be approximately 36.03 grams.
Step-by-step explanation:
In the balanced chemical equation 2MnO₂ + 4KOH + O₂ → 2K₂MnO₄ + 2H₂O, the stoichiometry implies that 2 moles of MnO₂ react with 4 moles of KOH to produce 2 moles of H₂O. First, determine the moles of each reactant given: MnO₂ = 166.0 g / (86.94 g/mol) ≈ 1.907 moles, KOH = 222.1 g / (56.11 g/mol) ≈ 3.959 moles, and O₂ = 28.15 g / (32.00 g/mol) ≈ 0.8809 moles.
Comparing the moles of reactants, it's evident that KOH is the limiting reactant, as it is less than the required stoichiometric ratio (4 moles). Therefore, 3.959 moles of KOH will produce 2 moles of H₂O. Now, calculate the molar mass of H₂O (18.02 g/mol) and find the mass produced: 2 moles * 18.02 g/mol ≈ 36.03 grams of H₂O formed.
The calculation reveals that with the given quantities of MnO₂, KOH, and O₂, the mass of H₂O formed will be approximately 36.03 grams, as limited by the amount of KOH available in the reaction. The stoichiometry of the balanced equation allows us to determine the limiting reactant and, consequently, the maximum amount of product formed.