When heated, lithium reacts with nitrogen to form lithium nitride:
6Li+N2 --> 2Li3N
What is the theoretical yield of Li3N in grams when 12.4 g of Li is heated with 33.9g of N2
This is a limiting reagent problem. How do I know that? Because BOTH reactants are given, not just one.
1. You have the equation and it is balanced.
2a. Convert 12.4 g Li to moles. moles = grams/molar mass.
2b. Convert 33.9 g N2 to moles the same way.
3a. Using the coefficients in the balanced equation, convert moles Li in 2a to moles Li3N.
3a. Same process, convert moles N2 in 2b to moles Li3N.
3c. It is more than likely that the two answers for moles Li3N will be different which means one of your values is wrong. In limiting reagent problems the correct answer is ALWAYS the smaller value and the reagent producing that smaller value is the limiting reagent.
4. Now convert moles Li3N from 3c to grams. g = moles x molar mass. This is the theoretical yield.
Post your work if you get stuck.
To calculate the theoretical yield of Li3N in grams, you need to determine the limiting reactant and use stoichiometry.
First, we need to find the molar masses of Li and N2:
- The molar mass of Li is approximately 6.94 g/mol.
- The molar mass of N2 is approximately 28.02 g/mol.
Next, we determine the moles of Li and N2:
- The moles of Li can be calculated by dividing the mass of Li (12.4 g) by its molar mass: 12.4 g / 6.94 g/mol ≈ 1.78 mol.
- The moles of N2 can be calculated by dividing the mass of N2 (33.9 g) by its molar mass: 33.9 g / 28.02 g/mol ≈ 1.21 mol.
Now, we need to determine the stoichiometric ratio between Li and Li3N based on the balanced equation:
- According to the balanced equation, 6 moles of Li react with 1 mole of N2 to produce 2 moles of Li3N.
Next, we need to determine the limiting reactant:
- To find the limiting reactant, compare the moles of Li and N2. In this case, Li is in excess, and N2 is the limiting reactant because it produces fewer moles of Li3N.
- The stoichiometric ratio tells us that 1 mole of N2 produces 2 moles of Li3N.
Now, we can calculate the moles of Li3N produced:
- Since the moles of N2 (1.21 mol) are limiting, we can multiply it by the stoichiometric ratio (2 moles of Li3N/1 mole of N2) to find the moles of Li3N produced: 1.21 mol × (2 moles of Li3N/1 mole of N2) = 2.42 mol.
Finally, we can calculate the theoretical yield of Li3N in grams:
- The molar mass of Li3N is approximately 34.83 g/mol.
- Multiply the moles of Li3N (2.42 mol) by its molar mass: 2.42 mol × 34.83 g/mol ≈ 84.25 g.
Therefore, the theoretical yield of Li3N is approximately 84.25 grams when 12.4 g of Li is heated with 33.9 g of N2.