Lithium and nitrogen react in a combination reaction to produce lithium nitride:
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.
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.
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.
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.
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.
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.
To find the theoretical yield of lithium nitride, we need to determine the limiting reagent first. The limiting reagent is the reactant that is completely consumed and determines the maximum amount of product that can be formed.
To find the limiting reagent, we can calculate how many moles of lithium and nitrogen we have in the 3.00 g of each reagent.
The molar mass of lithium (Li) is 6.94 g/mol, and the molar mass of nitrogen (N2) is 28.02 g/mol.
Moles of lithium (Li):
moles = mass / molar mass
moles = 3.00 g / 6.94 g/mol
moles = 0.432 mol
Moles of nitrogen (N2):
moles = mass / molar mass
moles = 3.00 g / 28.02 g/mol
moles = 0.107 mol
Now, we need to compare the mole ratio of the reactants to determine which one is the limiting reagent. The balanced equation shows that the ratio of Li to N2 is 6:1. This means that for every 6 moles of Li, we need 1 mole of N2.
Since we have only 0.107 mol of N2, which is significantly less than the required ratio, nitrogen is the limiting reagent.
Next, we need to calculate the moles of lithium nitride formed using the stoichiometry of the balanced equation:
From the balanced equation: 6 Li + N2 → 2 Li3N
Here, 1 mole of N2 reacts with 2 moles of Li3N.
Moles of Li3N formed = moles of N2 * (2 moles of Li3N / 1 mole of N2)
Moles of Li3N formed = 0.107 mol * (2 mol / 1 mol)
Moles of Li3N formed = 0.214 mol
Finally, we can calculate the mass of lithium nitride formed using its molar mass:
Mass of Li3N = moles of Li3N * molar mass of Li3N
Mass of Li3N = 0.214 mol * (49.94 g/mol + 3 * 6.94 g/mol)
Mass of Li3N = 0.214 mol * 73.7 g/mol
Mass of Li3N = 15.77 g
Therefore, the theoretical yield of lithium nitride is 15.77 g.
To determine the theoretical yield of lithium nitride (Li3N) in this reaction, we need to calculate the amount of limiting reagent. The limiting reagent is the one that is completely consumed and determines the maximum amount of product that can be produced.
1. Start by determining the number of moles of each reagent:
- Lithium (Li): Given mass = 3.00 g
molar mass of Li = 6.94 g/mol
moles of Li = given mass / molar mass of Li
- Nitrogen (N2): Given mass = 3.00 g
molar mass of N2 = 28.00 g/mol
moles of N2 = given mass / molar mass of N2
2. Use the balanced equation to determine the stoichiometry between the reagents and the product:
According to the balanced equation: 6Li(s) + N2(g) → 2Li3N(s)
The ratio of moles between Li and Li3N is 6:2 (or 3:1)
The ratio of moles between N2 and Li3N is 1:2
3. Compare the moles of each reagent to determine the limiting reagent:
The limiting reagent is the one that produces fewer moles of the product. Compare the moles of Li and N2 to find the limiting reagent.
4. Calculate the moles of Li3N formed from the limiting reagent:
Since the balanced equation shows a 1:2 stoichiometric ratio between N2 and Li3N, the moles of Li3N produced will be twice the moles of N2.
5. Convert moles of Li3N to grams:
Multiply the moles of Li3N by its molar mass (lithium nitride).
The resulting value will be the theoretical yield of lithium nitride (Li3N) in grams.