To determine how many grams of nitrogen are required to produce 0.65 moles of lithium nitride (Li₃N) with lithium in excess, follow these steps:
1. Write the balanced chemical equation:
[tex]\[
6 \text{Li} + \text{N}_2 \rightarrow 2 \text{Li}_3\text{N}
\][/tex]
2. Interpret the balanced equation:
- The equation shows that 1 mole of nitrogen gas (N₂) produces 2 moles of lithium nitride (Li₃N).
3. Relate moles of N₂ to moles of Li₃N:
- Since 1 mole of N₂ produces 2 moles of Li₃N, producing 1 mole of Li₃N requires 0.5 moles of N₂.
4. Calculate the moles of nitrogen required for 0.65 moles of Li₃N:
[tex]\[
\text{Moles of N}_2 = 0.5 \times \text{moles of Li}_3\text{N}
\][/tex]
[tex]\[
\text{Moles of N}_2 = 0.5 \times 0.65 = 0.325 \text{ moles of N}_2
\][/tex]
5. Determine the mass of nitrogen needed:
- The molar mass of nitrogen gas (N₂) can be calculated by adding the atomic masses of two nitrogen atoms. Each nitrogen atom has an atomic mass of approximately 14 g/mol.
[tex]\[
\text{Molar mass of N}_2 = 14 \text{ g/mol} \times 2 = 28 \text{ g/mol}
\][/tex]
- Now, calculate the mass of 0.325 moles of N₂:
[tex]\[
\text{Mass of N}_2 = \text{moles of N}_2 \times \text{molar mass of N}_2
\][/tex]
[tex]\[
\text{Mass of N}_2 = 0.325 \times 28 = 9.1 \text{ grams}
\][/tex]
Therefore, to produce 0.65 moles of lithium nitride, you need 0.325 moles of nitrogen gas, which corresponds to 9.1 grams of nitrogen.
