Final answer:
No, 2.39 g of lithium nitride will not react completely with 2.90 g of water.
Step-by-step explanation:
In order to determine whether 2.39 g of lithium nitride will react completely with 2.90 g of water, we need to compare the amounts of reactants to the stoichiometry of the reaction. The balanced chemical equation for the reaction is:
Li3N + 3H2O → 3LiOH + NH3
From the equation, we can see that 1 mole of Li3N reacts with 3 moles of H2O. To find the limiting reactant, we need to calculate the moles of each reactant:
Lithium nitride (Li3N):
Molar mass of Li3N = 34.83 g/mol
Moles of Li3N = (2.39 g Li3N) / (34.83 g/mol Li3N) = 0.0686 mol Li3N
Water (H2O):
Molar mass of H2O = 18.02 g/mol
Moles of H2O = (2.90 g H2O) / (18.02 g/mol H2O) = 0.1611 mol H2O
Since the stoichiometry of the reaction requires 3 moles of H2O per mole of Li3N, we can see that 0.1611 mol H2O is in excess and will not react completely. Therefore, the answer is No, 2.39 g of lithium nitride will not react completely with 2.90 g of water.