Haber-Bosch: Limiting Reagent & Theoretical Yield

The Haber–Bosch synthesis converts nitrogen gas to ammonia, enabling fertilizer production and supporting agricultural productivity.
Balanced equation (stoichiometry):
\mathrm{N2} + 3\,\mathrm{H2} \rightarrow 2\,\mathrm{NH_3}
Key concepts: balancing equations, identifying limiting reactant, calculating theoretical yield, and determining leftover reactants.

Initial amounts: 4\ \text{moles of } \mathrm{N2} \quad\text{and}\quad 9\ \text{moles of } \mathrm{H2}
Assumed product: \mathrm{NH_3} (ammonia) is the product; this makes the problem a limiting-reagent calculation.
The transcript emphasizes checking whether the equation is balanced and recognizing limiting reagents when two reactants are given.

Determine how much NH3 could be formed from each reactant if it were used completely:
- From (\mathrm{N2}): 4\ \text{mol }\mathrm{N2} \times \frac{2\ \text{mol }\mathrm{NH3}}{1\ \text{mol }\mathrm{N2}} = 8\ \text{mol }\mathrm{NH_3}
- From (\mathrm{H2}): 9\ \text{mol }\mathrm{H2} \times \frac{2\ \text{mol }\mathrm{NH3}}{3\ \text{mol }\mathrm{H2}} = 6\ \text{mol }\mathrm{NH_3}
Conclusion: The limiting reactant is (\mathrm{H_2}); the theoretical maximum NH3 is 6 moles.
Transcript note: the presenter states this explicitly and identifies (\mathrm{H_2}) as the limiting reagent.

Theoretical yield of NH3 (from limiting reactant): 6\ \text{mol } \mathrm{NH_3}
Amount of (\mathrm{H_2}) consumed to produce 6 mol NH3:
- Stoichiometry shows that 2 NH3 require 3 H2, so for 6 NH3:
  6\ \text{mol } \mathrm{NH3} \times \frac{3\ \text{mol }\mathrm{H2}}{2\ \text{mol }\mathrm{NH3}} = 9\ \text{mol }\mathrm{H2}
Amount of (\mathrm{N_2}) consumed to produce 6 NH3:
- 2 NH3 require 1 N2, so for 6 NH3:
  6\ \text{mol } \mathrm{NH3} \times \frac{1\ \text{mol }\mathrm{N2}}{2\ \text{mol }\mathrm{NH3}} = 3\ \text{mol }\mathrm{N2}
Therefore, after the reaction, limiting reactant ( (\mathrm{H2}) ) is completely consumed, and ( \mathrm{N2} ) remains in excess.

Leftover amounts:
- (\mathrm{N2}) consumed: 3 mol; initial (\mathrm{N2}) = 4 mol \rightarrow leftover (\mathrm{N_2} = 4 - 3 = 1) mol
- (\mathrm{H2}) consumed: 9 mol; initial (\mathrm{H2} = 9) mol \rightarrow leftover (\mathrm{H_2} = 0) mol
Transcript note and correction:
- The transcript claims the leftover (\mathrm{N2} = 4) mol, which is incorrect. The correct leftover is 1 mol of (\mathrm{N2}).
Summary of results for the given amounts:
- Theoretical yield of NH3: 6\ \text{mol}
- Amount of (\mathrm{H_2}) consumed: 9\ \text{mol}
- Amount of (\mathrm{N_2}) consumed: 3\ \text{mol}
- Leftover reagents: (\mathrm{N2} = 1) mol, (\mathrm{H2} = 0) mol

Key concept: the limiting reactant controls the amount of product; excess reactants remain after the reaction.
Coefficients as multipliers: the coefficients in the balanced equation tell you how many moles of each substance participate per a given amount of reaction.
Common pitfall (illustrated in the car analogy): it’s easy to misinterpret which reagent is limiting or to miscalculate leftover amounts if you don’t consistently apply the stoichiometric ratios.
The teacher uses analogies (cars with chassis and wheels) to illustrate how the ratio dictates how many products can be formed given available inputs.

Car analogy: one car requires 1 chassis and 4 wheels; scaling up to multiple cars uses the same ratio. For example, with 4 cars, you’d need 4 chassis and 16 wheels; this illustrates how ratios drive product amounts.
Reiterated idea: practice with stoichiometry involves setting up conversion factors from the balanced equation and tracking units to ensure proper cancellation.
Practice resources mentioned:
- Canvas practice quizzes (current assignments tab) covering the material.
- Practice exam one available at the bottom of the page.
- ALEKS extra practice in the ebook.
- Recommendation: reread the material and complete quizzes and the exam to prepare.

Always start with a balanced equation:
- \mathrm{N2} + 3\,\mathrm{H2} \rightarrow 2\,\mathrm{NH_3}
For each reactant, compute the amount of product it could form if it were the limiting reactant to identify the true limiting reagent.
The theoretical yield equals the amount of product formed using the limiting reagent.
Use stoichiometric ratios to determine how much of the other reactants are consumed and how much remains.
Check units carefully to ensure proper cancellation throughout the calculation.

Practical importance: Ammonia production is central to fertilizer manufacturing, enabling large-scale agriculture and food production.
Broader implications (not deeply discussed in the transcript but relevant): environmental considerations (e.g., energy use, emissions) associated with Haber-Bosch processes; efficiency and yield optimization are ongoing industrial concerns.

For the reaction \mathrm{N2} + 3\,\mathrm{H2} \rightarrow 2\,\mathrm{NH3} with 4 mol (\mathrm{N2} ) and 9 mol (\mathrm{H_2} ):
- The limiting reagent is (\mathrm{H_2}).
- The theoretical yield of NH3 is 6\ \text{mol}.
- (\mathrm{H2} ) is completely consumed (0 leftover), and (\mathrm{N2} ) left over is 1\ \text{mol}.
Always cross-check calculations; the limiting reagent determines the maximum product, and any excess is left unused.

Reaction: \mathrm{N2} + 3\,\mathrm{H2} \rightarrow 2\,\mathrm{NH_3}
NH3 from N2: 4\ \text{mol } \mathrm{N2} \times \frac{2\ \text{mol } \mathrm{NH3}}{1\ \text{mol } \mathrm{N2}} = 8\ \text{mol } \mathrm{NH3}
NH3 from H2: 9\ \text{mol } \mathrm{H2} \times \frac{2\ \text{mol } \mathrm{NH3}}{3\ \text{mol } \mathrm{H2}} = 6\ \text{mol } \mathrm{NH3}
Theoretical NH3 yield: 6\ \text{mol } \mathrm{NH_3}
H2 consumed: 9\ \text{mol } \mathrm{H_2}
N2 consumed: 3\ \text{mol } \mathrm{N_2}
N2 leftover: 4 - 3 = 1\ \text{mol } \mathrm{N_2}
H2 leftover: 9 - 9 = 0\ \text{mol } \mathrm{H_2}