Final answer:
To react with 1.50 mol of Zn in the given chemical equation, one would need 3.00 mol of MnO₂ and 1.50 mol of H₂O.
Step-by-step explanation:
The question involves using stoichiometry to determine the amount of MnO₂ and H₂O that will completely react with a given amount of zinc (Zn) in the specified chemical reaction. The balanced chemical equation for this reaction is already provided:
Zn(s) + 2MnO₂(s) + H₂O(l) → Zn(OH)₂(aq) + Mn₂O₃(s)
Given that the stoichiometry of the reaction requires two moles of MnO₂ and one mole of H₂O for every mole of Zn reacted, the calculation for the amount of reactants needed for 1.50 mol Zn is straightforward.
To react completely with 1.50 mol of Zn, you would need:
- 2 moles of MnO₂ for every 1 mole of Zn, which calculates to 1.50 mol Zn × 2 mol MnO₂/mol Zn = 3.00 mol MnO₂.
- 1 mole of H₂O for every mole of Zn, which equals to 1.50 mol H₂O.
Therefore, to react with 1.50 mol of Zn, one would require 3.00 mol of MnO₂ and 1.50 mol of H₂O.