Questions: When heated, lithium reacts with nitrogen to form lithium nitride: 6 Li(s) + N2(g) -> 2 Li3 N(s) Part: 0 / 2 Part 1 of 2 What is the theoretical yield of Li3 N in grams when 12.8 g of Li is heated with 34.9 g of N2? Be sure your answer has the correct number of significant digits. g

When heated, lithium reacts with nitrogen to form lithium nitride:
6 Li(s) + N2(g) -> 2 Li3 N(s)

Part: 0 / 2

Part 1 of 2

What is the theoretical yield of Li3 N in grams when 12.8 g of Li is heated with 34.9 g of N2? Be sure your answer has the correct number of significant digits.
g

Transcript text: When heated, lithium reacts with nitrogen to form lithium nitride: \[ 6 \mathrm{Li}(s)+\mathrm{N}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{Li}_{3} \mathrm{~N}(s) \] Part: $0 / 2$ Part 1 of 2 What is the theoretical yield of $\mathrm{Li}_{3} \mathrm{~N}$ in grams when 12.8 g of Li is heated with 34.9 g of $\mathrm{N}_{2}$ ? Be sure your answer has the correct number of significa digits. $\square \mathrm{g}$

Solution

Solution Steps

Step 1: Determine the Molar Masses

First, we need to determine the molar masses of lithium (Li), nitrogen ($\mathrm{N}_2$), and lithium nitride ($\mathrm{Li}_3\mathrm{N}$).

Molar mass of Li: $6.941 \, \text{g/mol}$
Molar mass of $\mathrm{N}_2$: $2 \times 14.007 = 28.014 \, \text{g/mol}$
Molar mass of $\mathrm{Li}_3\mathrm{N}$: $3 \times 6.941 + 14.007 = 34.829 \, \text{g/mol}$

Step 2: Convert Masses to Moles

Convert the given masses of Li and $\mathrm{N}_2$ to moles.

Moles of Li: \[ \frac{12.8 \, \text{g}}{6.941 \, \text{g/mol}} = 1.843 \, \text{mol} \]
Moles of $\mathrm{N}_2$: \[ \frac{34.9 \, \text{g}}{28.014 \, \text{g/mol}} = 1.246 \, \text{mol} \]

Step 3: Determine the Limiting Reactant

The balanced chemical equation is: \[ 6 \mathrm{Li}(s) + \mathrm{N}_2(\mathrm{~g}) \rightarrow 2 \mathrm{Li}_3 \mathrm{~N}(s) \]

From the stoichiometry of the reaction:

6 moles of Li react with 1 mole of $\mathrm{N}_2$.

Calculate the required moles of $\mathrm{N}_2$ for the available moles of Li: \[ \frac{1.843 \, \text{mol Li}}{6} = 0.3072 \, \text{mol} \, \mathrm{N}_2 \]

Since 0.3072 mol of $\mathrm{N}_2$ is less than the available 1.246 mol, Li is the limiting reactant.

Step 4: Calculate Theoretical Yield

Using the limiting reactant (Li), calculate the moles of $\mathrm{Li}_3\mathrm{N}$ produced: \[ \frac{1.843 \, \text{mol Li}}{6} \times 2 = 0.6143 \, \text{mol} \, \mathrm{Li}_3\mathrm{N} \]

Convert moles of $\mathrm{Li}_3\mathrm{N}$ to grams: \[ 0.6143 \, \text{mol} \times 34.829 \, \text{g/mol} = 21.41 \, \text{g} \]