Consider the reaction of 2.5 grams of Li (s) reacting with 0.5 grams of N2 (g) to produce Li3N (s). A) How many total grams of Li3 N (s) would be produced? B) Which of the starting elements is the limiting reagent, and how many grams of the non-limiting reagent remain after the reaction has completed?

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.

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