Lithium and nitrogen react in a combination reaction to produce lithium nitride: 6Li(s) + N_2(g) → 2Li_3N(s)

In a particular experiment, 1 00 g samples of each reagent are reacted. The theoretical yield of Iithium nitride is .

a. 5.0
b. 0.84
c. 1.01
d. 1.67
e. 2.50

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 N₂ 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 N₂.

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 N₂ out of 1 g given is required to react completely with 1 g of Li.

Therefore, Li is the limiting reactant and N₂ 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.

i think   the answer is c hope i'm right