Final answer:
2.39 g of lithium nitride will not react completely with 2.90 g of water because the masses are not stoichiometrically balanced for a complete reaction; water is the limiting reactant.
Step-by-step explanation:
To determine if 2.39 g of lithium nitride, Li₃N(s), will react completely with 2.90 g of water, we must consider the stoichiometry of the reaction. Lithium nitride reacts with water to produce lithium hydroxide and ammonia gas:
Li₃N(s) + 3H₂O(l) → 3LiOH(aq) + NH₃(g)
First, let's calculate the moles of lithium nitride and water:
- Molar mass of Li₃N = 3(6.939 g/mol) + 14.007 g/mol = 34.824 g/mol.
- Moles of Li₃N = 2.39 g / 34.824 g/mol = 0.0686 mol.
- Molar mass of H₂O = 18.015 g/mol.
- Moles of H₂O = 2.90 g / 18.015 g/mol = 0.161 mol.
According to the balanced chemical equation, 1 mole of Li₃N reacts with 3 moles of H₂O. So, for every mole of Li₃N, we need three times more moles of H₂O. However, we have less than three times the moles of water for the moles of lithium nitride available (0.0686 mol Li₃N and 0.161 mol H₂O).
Therefore, water is the limiting reactant, and the reaction will not go to completion with the given amounts. The correct answer is D) No, because the masses are not stoichiometrically balanced for a complete reaction.