Question

Which is the limiting reagent in the reaction 3Mg + N2 → Mg3N2?

Answer 1

To determine the limiting reagent in the reaction \( 3\text{Mg} + \text{N}_2 \rightarrow \text{Mg}_3\text{N}_2 \), we need to compare the mole ratio of the reactants with the given quantities of each reactant.

From the balanced equation:
- 3 moles of magnesium (Mg) react with 1 mole of nitrogen gas (N2).

1. **Identify the starting amounts**: Determine how much of each reactant you have. Let's say you have:
- \( x \) moles of Mg
- \( y \) moles of N2

2. **Determine the stoichiometric needs**: For every 3 moles of Mg, you need 1 mole of N2.

3. **Set up the ratios**:
- For magnesium, the required moles of N2 would be \( \frac{x}{3} \).
- For nitrogen, the required moles of Mg would be \( 3y \).

4. **Compare the available amounts**:
- If you have enough N2 to react with your available Mg, then Mg is in excess, and N2 is the limiting reagent.
- Conversely, if you have enough Mg to consume the available N2, then N2 is in excess, and Mg is the limiting reagent.

5. **Conclusion**: Whichever reactant you run out of first is the limiting reagent.

**Example**: If you have 6 moles of Mg and 1 mole of N2:
- For 6 moles of Mg: \( \frac{6 \, \text{moles of Mg}}{3} = 2 \, \text{moles of N2 needed} \)
- You only have 1 mole of N2, which is not enough.

In this case, N2 is the limiting reagent, and Mg is in excess.

If you provide specific amounts for Mg and N2, I can give you the exact limiting reagent based on those quantities.