Answer:
A) The amount in grams of Li₃N produced is 1.243 g
B) N₂, is the limiting reagent
The mass of the non-limiting reagent, Li, remaining after the reaction is completed is 1.757 g
Explanation:
The given parameters are;
The mass of Li(s) = 2.5 grams
The mass of N₂ (g) = 0.5 grams
The chemical equation for the reaction can be presented as follows;
6 Li (s) + N₂ (g) → 2 Li₃N
Therefore, 6 moles of Li reacts with 1 mole of N₂ to produce 2 moles of Li₃N
The molar mass of Li = 6.941 g/mol
The molar mass of N₂ = 28.0134 g/mol
The number of moles of a reactant or product, n is given by the relation;
n = Mass of substance/(Molar mass of the substance)
For lithium, Li, n = 2.5/6.941 = 0.3602 moles
For Nitrogen gas, N₂, n = 0.5/28.0134 = 0.01785 moles
A) Given that 1 mole of N₂ to produces 2 moles of Li₃N
0.01785 moles of N₂ will produces 2×0.01785 = 0.0357 moles of Li₃N
The molar mass of Li₃N = 34.83 g/mol
The mass of Li₃N = 34.83 g/mol × 0.0357 moles = 1.243 g
B) 6 moles of Li reacts with 1 mole of N₂ to produce 2 moles of Li₃N
0.3602 moles will reacts with 1/6×0.3602 = 0.06003 mole of N₂
Therefore, N₂, is the limiting reagent and we have;
0.01785 moles of N₂ will react with 6×0.01785 = 0.1071 moles of Li
The number of of moles of Li left = 0.3602 - 0.1071 =0.2531 moles
The mass of lithium left = 0.2531 moles × 6.941 g/mol = 1.757 g
The mass of lithium remaining after the reaction is completed = 1.757 g.
