Question

In a particular experiment, 2.50-g samples of each reagent are reacted. The theoretical yield of lithium nitride is ________ g. Molar mass of Li is 6.94 g/mol. Molar mass of N2 is 28.02 g/mol. Molar mass of Li3N is 34.83 g/mol.

Answer 1

Answer:

4.18 g

Explanation:

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Given: For Li

Given mass = 2.50 g

Molar mass of Li  = 6.94 g/mol

Moles of Li  = 2.50 g / 6.94 g/mol = 0.3602 moles

Given: For $N_2$

Given mass = 2.50 g

Molar mass of $N_2$ = 28.02 g/mol

Moles of $N_2$ = 2.50 g / 28.02 g/mol = 0.08924 moles

According to the given reaction:

$6Li+N_2\rightarrow 2Li_3N$

6 moles of Li react with 1 mole of $N_2$

1 mole of Li react with 1/6 mole of $N_2$

0.3602 mole of Li react with $\frac {1}{6}\times 0.3602$ mole of $N_2$

Moles of $N_2$ that will react = 0.06 moles

Available moles of $N_2$ = 0.08924 moles

$N_2$ is in large excess. (0.08924 > 0.06)

Limiting reagent is the one which is present in small amount. Thus,

Li is limiting reagent.

The formation of the product is governed by the limiting reagent. So,

6 moles of Li gives 2 mole of $Li_3N$

1 mole of Li gives 2/6 mole of $Li_3N$

0.3602 mole of Li react with $\frac {2}{6}\times 0.3602$ mole of $Li_3N$

Moles of $Li_3N$ = 0.12

Molar mass of $Li_3N$ = 34.83 g/mol

Mass of $Li_3N$ = Moles × Molar mass = 0.12 × 34.83 g = 4.18 g

Theoretical yield = 4.18 g