Concept:Stoichiometry allows us to scale up a reaction based on the fixed ratio established by the balanced chemical equation.
Formula:$$ \frac{\text{Moles of N}_2}{\text{Coefficient of N}_2} = \frac{\text{Moles of NH}_3}{\text{Coefficient of NH}_3} $$
Solution:- The balanced equation (\( \text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3 \)) explicitly states that 1 mole of \( \text{N}_2 \) produces 2 moles of \( \text{NH}_3 \).
- This is a straightforward 1:2 ratio. You will always produce exactly double the moles of ammonia relative to the nitrogen consumed.
- We are provided with 2.5 moles of \( \text{N}_2 \).
- Therefore, Moles of \( \text{NH}_3 = 2.5 \times 2 = 5.0 \text{ moles} \).
Why other options are incorrect:- 2.5 moles: Assumes a 1:1 ratio, completely ignoring the stoichiometric coefficients.
- 2 moles: This is just the stoichiometric coefficient for ammonia, ignoring the 2.5 mole input constraint.
- 7.5 moles: Assumes a 1:3 ratio, confusing the hydrogen coefficient with the ammonia coefficient.
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