Answer:
The maximum amount of Li3N that can be formed is 0.663 mol Li3N.
Answer: 0.663 mol
Explanation:
To find the maximum amount of Li3N that can be formed from the given amounts of Li and N2, we need to determine the limiting reactant.
1. First, we need to calculate the number of moles of Li and N2.
Given:
Mass of Li = 13.82 g
Molar mass of Li = 6.94 g/mol
Number of moles of Li = Mass of Li / Molar mass of Li = 13.82 g / 6.94 g/mol = 1.99 mol
Number of moles of N2 = 0.338 mol (given)
2. Next, we need to determine the stoichiometric ratio between Li and Li3N from the balanced chemical equation.
The balanced equation is:
6Li + N2 → 2Li3N
From the balanced equation, we can see that 6 moles of Li react with 1 mole of N2 to produce 2 moles of Li3N.
3. Now, we compare the number of moles of Li and N2 to determine the limiting reactant.
Using the stoichiometric ratio, we can determine the maximum amount of Li3N that can be formed from the limiting reactant.
a. For Li: The stoichiometric ratio is 6 moles of Li to 2 moles of Li3N.
So, the maximum moles of Li3N formed from Li = (1.99 mol Li) * (2 mol Li3N / 6 mol Li) = 0.663 mol Li3N
b. For N2: The stoichiometric ratio is 1 mole of N2 to 2 moles of Li3N.
So, the maximum moles of Li3N formed from N2 = (0.338 mol N2) * (2 mol Li3N / 1 mol N2) = 0.676 mol Li3N
4. Finally, we compare the amounts of Li3N formed from each reactant, and the lower value is the maximum amount that can be formed.
Therefore, the maximum amount of Li3N that can be formed is 0.663 mol Li3N.
Answer: 0.663 mol
