Answer:
Option D. 1.67 g
Explanation:
The balanced equation for the reaction is given below:
6Li (s) + N₂ (g) → 2Li₃N (s)
Next, we shall determine the masses of Li and Nâ‚‚ that reacted and the mass of
Li₃N produced from the balanced equation. This can be obtained as follow:
Molar mass of Li = 7 g/mol
Mass of Li from the balanced equation = 6 × 7 = 42 g
Molar mass of N₂ = 14 × 2 = 28 g/mol
Mass of N₂ from the balanced equation = 1 × 28 = 28 g
Molar mass of Li₃N = (7×3) + 14 = 21 + 14 = 35 g/mol
Mass of Li₃N from the balanced equation = 2 × 35 = 70 g
SUMMARY:
From the balanced equation above,
42 g of Li reacted with 28 g of N₂ to produce 70 g of Li₃N.
Next, we shall determine the limiting reactant. This can be obtained as illustrated below:
From the balanced equation above,
42 g of Li reacted with 28 g of Nâ‚‚.
Therefore, 1 g of Li will react with =
(1 × 28)/42 = 0.67 g of N₂
From the calculation made above, we can see that only 0.67 g of Nâ‚‚ out of 1 g given is required to react completely with 1 g of Li.
Therefore, Li is the limiting reactant and Nâ‚‚ is the excess reactant.
Finally, we shall determine the theoretical yield of Iithium nitride, Li₃N.
The theoretical yield of Iithium nitride, Li₃N, can be obtained by using the limiting reactant as illustrated below:
From the balanced equation above,
42 g of Li reacted to produce 70 g of Li₃N.
Therefore, 1 g of Li will react to produce = (1 × 70)/42 = 1.67 g of Li₃N.
Thus, the theoretical yield of Iithium nitride, Li₃N is 1.67 g.
