Lithium and nitrogen react in a combination reaction to produce lithium nitride:

Question

6Li(s)+N2(g)→2Li3N(s)


In a particular experiment, 3.00-g samples of each reagent are reacted. The theoretical yield of lithium nitride is __________ g.

Answer 1

To calculate the theoretical yield of lithium nitride, we need to determine the limiting reactant first. The limiting reactant is the reactant that is completely consumed, thus determining the maximum amount of product that can be formed.

To determine the limiting reactant, we need to compare the number of moles of each reactant.

Molar mass of Li = 6.94 g/mol
Molar mass of N2 = 28.02 g/mol

Moles of Li = (mass of Li / molar mass of Li) = (3.00 g / 6.94 g/mol) = 0.432 mol
Moles of N2 = (mass of N2 / molar mass of N2) = (3.00 g / 28.02 g/mol) = 0.107 mol

The ratio between Li and Li3N in the balanced equation is 6:2, which simplifies to 3:1. This means that for each mole of Li, we need 1/3 mole of Li3N.

Considering this ratio, the moles of Li3N that can be formed by the available moles of Li is:
0.432 mol Li × (1/3 mol Li3N) = 0.144 mol Li3N

The moles of Li3N that can be formed by the available moles of N2 is:
0.107 mol N2 × (1 mole Li3N / 1 mole N2) = 0.107 mol Li3N

We can see that the moles of Li3N formed by N2 are smaller than the moles of Li3N formed by Li, meaning that N2 is the limiting reactant.

To find the theoretical yield of Li3N, we need to multiply the moles of Li3N formed by the molar mass of Li3N:
Theoretical yield of Li3N = (moles of Li3N) × (molar mass of Li3N)
Theoretical yield of Li3N = (0.107 mol) × (34.83 g/mol) = 3.73 g

Therefore, the theoretical yield of lithium nitride is 3.73 g.

Answer 2

Hey Bot. No, moles Li3N formed by 3 g N2 is GREATER than moles formed by 3 g Li; therefore, Li is the limiting reagent. I think you should rethink and recalculate.

Answer 3

Apologies for the mistake. You are correct.

To determine the limiting reactant, we need to compare the number of moles of each reactant.

Moles of Li = (mass of Li / molar mass of Li) = (3.00 g / 6.94 g/mol) = 0.432 mol
Moles of N2 = (mass of N2 / molar mass of N2) = (3.00 g / 28.02 g/mol) = 0.107 mol

The ratio between Li and Li3N in the balanced equation is 6:2, which simplifies to 3:1. This means that for each mole of Li, we need 1/3 mole of Li3N.

Considering this ratio, the moles of Li3N that can be formed by the available moles of Li is:
0.432 mol Li × (1/3 mol Li3N/1 mol Li) = 0.144 mol Li3N

The moles of Li3N that can be formed by the available moles of N2 is:
0.107 mol N2 × (1 mole Li3N / 1 mole N2) = 0.107 mol Li3N

We can see that the moles of Li3N formed by Li are greater than the moles of Li3N formed by N2, meaning that Li is the limiting reactant.

To find the theoretical yield of Li3N, we need to multiply the moles of Li3N formed by the molar mass of Li3N:
Theoretical yield of Li3N = (moles of Li3N) × (molar mass of Li3N)
Theoretical yield of Li3N = (0.144 mol) × (34.83 g/mol) = 5.01 g

Therefore, the theoretical yield of lithium nitride is 5.01 g.

Answer 4

Bot, you're wrong again. moles Li3N formed from 0.432 mols Li follows:
0.432 mol Li × (1/3 mol Li3N/1 mol Li) = 0.144 mol Li3N

This next step is your error which I've corrected.
moles Li3N formed from 0.107 moles N2 follows:
0.107 mol N2 × (2 mole Li3N / 1 mole N2) = 0.214 mol Li3N
Since the moles Li3N from Li is smaller than that formed from the moles N2 then Li must be the limiting reagent.
The remainder of the problem must be recalculated.

Answer 5

Apologies for the errors in my previous responses. Let's recalculate.

Moles of Li = (mass of Li / molar mass of Li) = (3.00 g / 6.94 g/mol) = 0.432 mol
Moles of N2 = (mass of N2 / molar mass of N2) = (3.00 g / 28.02 g/mol) = 0.107 mol

The ratio between Li and Li3N in the balanced equation is 6:2, which simplifies to 3:1. This means that for every 3 moles of Li, we need 1 mole of Li3N.

Moles of Li3N formed from 0.432 moles of Li follows:
0.432 mol Li × (1/3 mol Li3N / 1 mol Li) = 0.144 mol Li3N

Moles of Li3N formed from 0.107 moles of N2 follows:
0.107 mol N2 × (2 mol Li3N / 1 mol N2) = 0.214 mol Li3N

Since the moles of Li3N formed from Li (0.144 mol) are smaller than the moles of Li3N formed from N2 (0.214 mol), Li is indeed the limiting reagent.

To find the theoretical yield of Li3N, we need to multiply the moles of Li3N formed by the limiting reagent by the molar mass of Li3N.
Theoretical yield of Li3N = (moles of Li3N from limiting reagent) × (molar mass of Li3N)
Theoretical yield of Li3N = (0.144 mol) × (34.83 g/mol) = 5.01 g

Therefore, the theoretical yield of lithium nitride is 5.01 g.