Limiting Reagent

Question

In the reaction N₂ + 3H₂ → 2NH₃, if 28 g of N₂ is mixed with 8 g of H₂, the mass of NH₃ formed is: (N = 14, H = 1)

RankGuru · Accepted Answer

1. **Calculate Moles of Each Reactant**:
   - Moles of $N_2 = \frac{28}{28} = 1.0$ mol.
   - Moles of $H_2 = \frac{8}{2} = 4.0$ mol.
2. **Determine the Limiting Reagent**: From the balanced equation, 1 mol $N_2$ requires 3 mol $H_2$.
   - Available: 1.0 mol $N_2$ needs $1.0 	imes 3 = 3.0$ mol $H_2$.
   - We have 4.0 mol $H_2$, which is more than enough.
   - **$N_2$ is the limiting reagent** (completely consumed first).
3. **Calculate Product**:
   - 1 mol $N_2$ produces 2 mol $NH_3$.
   - Mass of $NH_3 = 2 	imes 17 = 34$ g.
4. **Check Excess H₂ Remaining**:
   - $H_2$ used = 3.0 mol, remaining = $4.0 - 3.0 = 1.0$ mol (i.e., 2 g excess).
5. **Takeaway**: Always compare the mole ratio of reactants to the stoichiometric ratio to identify the limiting reagent before computing the product mass.

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