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

The balanced chemical equation for the reaction is:

6 Li + N2 → 2 Li3N

From the equation, we can see that 6 moles of Li react with 1 mole of N2 to produce 2 moles of Li3N. We can use this information and the molar masses of the elements to calculate the theoretical yield of Li3N:

Molar mass of Li: 6.94 g/mol

Molar mass of N2: 28.02 g/mol

Molar mass of Li3N: 34.83 g/mol

First, we need to calculate the number of moles of Li in the reaction:

2.50 g Li × (1 mol Li/6.94 g Li) = 0.360 mol Li

Next, we need to calculate the limiting reagent in the reaction. Since we have 0.360 moles of Li, and 1 mole of N2 is needed for every 6 moles of Li, we have:

0.360 mol Li × (1 mol N2/6 mol Li) = 0.0600 mol N2

Therefore, N2 is the limiting reagent in the reaction. We can now calculate the theoretical yield of Li3N:

0.0600 mol N2 × (2 mol Li3N/1 mol N2) × (34.83 g Li3N/1 mol Li3N) = 4.74 g Li3N

Therefore, the theoretical yield of Li3N is 4.74 g.

Answer 2

4.18 g

Step-by-step explanation:

The formula for the calculation of moles is shown below:

`moles = (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}* 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}* 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