Chapter 3: Limiting Reagents Lecture

Overview of Limiting Reagents

• Limiting Reagent (LR) is defined as the reactant in a chemical reaction that is used up first.
• The limiting reagent determines and limits the maximum amount of product that can be obtained from a specific reaction.
• Identifying the limiting reagent is a critical step in laboratory procedures to predict theoretical yield.
• An excess reagent is the reactant that remains after the limiting reagent has been fully consumed.
• Calculations are required to determine which reagent is limiting, as it is not always obvious from the starting masses or mole counts.

Identifying Limiting Reagents via Stoichiometry

• To figure out which reactant is the limiting reagent and which is the excess reagent, one must perform stoichiometric calculations for each reactant.
• The Procedure:
  1. Start with the given mass (grams) of the first reactant.
  2. Calculate how many moles of the product can be produced from that specific amount of reactant.
  3. Repeat the same process for the second reactant (and any subsequent reactants).
  4. Compare the results in terms of moles of product.
  5. The reactant that produces the smallest value (the least amount of product) is the limiting reagent.
  6. The reactant that yields the higher value is the excess reagent.
• Once the limiting reagent is used up, the reaction stops; therefore, having "tens of thousands" or "tons" of the excess reagent is irrelevant to producing more product.

Illustrated Example: Synthesis of Methanol

• Reaction Equation: Carbon monoxide reacts with hydrogen to produce methanol ($CH_3OH$).
• Scenario from Sketch:
  • Carbon Monoxide ($CO$): Represented as black spheres. In the example, there are $4\text{ moles}$ of $CO$.
  • Hydrogen ($H_2$): Represented as gray spheres. In the example, there are $6\text{ moles}$ of $H_2$.
• Reaction Outcome:
  • The product formed is methanol (highlighted in circles in the diagram).
  • $3\text{ moles}$ of methanol are produced.
  • After the reaction, some carbon monoxide spheres remain.
• Conclusion:
  • Hydrogen ($H_2$) is the limiting reagent because it was used up first and limited the production to $3\text{ moles}$.
  • Carbon monoxide ($CO$) is the excess reagent because it remained at the end of the reaction.

Comprehensive Practice Problem: Urea Preparation

• Background: Urea is prepared by reacting ammonia ($NH_3$) with carbon dioxide ($CO_2$) according to the following balanced equation:
  • $2NH_3 + CO_2 \rightarrow (NH_2)_2CO + H_2O$
• Given Data:
  • Mass of Ammonia: $637.2\text{ g}$ of $NH_3$
  • Mass of Carbon Dioxide: $1142\text{ g}$ of $CO_2$

Part A: Determining the Limiting Reagent

• Step 1: Calculate moles of urea from Ammonia ($NH_3$)
  • Convert grams of $NH_3$ to moles: $637.2\text{ g } NH_3 \times \frac{1\text{ mol } NH_3}{17.03\text{ g } NH_3}$.
  • Convert moles of $NH_3$ to moles of Urea using the stoichiometric coefficient ($2:1$ ratio): $\frac{1\text{ mol Urea}}{2\text{ mol } NH_3}$.
  • Calculation: $\frac{637.2}{17.03 \times 2} = 18.7\text{ mol Urea}$.
• Step 2: Calculate moles of urea from Carbon Dioxide ($CO_2$)
  • Convert grams of $CO_2$ to moles: $1142\text{ g } CO_2 \times \frac{1\text{ mol } CO_2}{44.01\text{ g } CO_2}$.
  • Convert moles of $CO_2$ to moles of Urea using the stoichiometric coefficient ($1:1$ ratio): $\frac{1\text{ mol Urea}}{1\text{ mol } CO_2}$.
  • Calculation: $\frac{1142}{44.01} = 25.95\text{ mol Urea}$.
• Conclusion for Part A:
  • Because $18.7\text{ moles}$ is smaller than $25.95\text{ moles}$, Ammonia ($NH_3$) is the limiting reagent.
  • Carbon dioxide ($CO_2$) is the excess reagent.
  • The maximum number of moles of urea that can be formed is $18.7\text{ mol}$.

Part B: Calculating the Mass of Urea Formed

• The mass produced is determined by the limiting reagent ($18.7\text{ mol Urea}$).
• Molar Mass of Urea calculation: Based on two nitrogens, four hydrogens, one carbon, and one oxygen, the molar mass provided is $66.06\text{ g/mol}$.
• Final Calculation:
  • $18.7\text{ mol Urea} \times 66.06\text{ g/mol} = 1235.3\text{ g Urea}$.
• Important Correction: There is a typo on the lecture PowerPoint which states $1124\text{ g}$, but the correct calculator input result is $1235.3\text{ g}$.

Part C: Calculating Amount of Excess Reagent Left

• To find the remaining excess reagent, you must first calculate how much of the excess reactant ($CO_2$) was actually used to produce the maximum amount of product.
• Step 1: Calculate $CO_2$ used
  • Start with the maximum moles of urea: $18.7\text{ mol Urea}$.
  • Use the mole ratio to find moles of $CO_2$ used: $18.7\text{ mol Urea} \times \frac{1\text{ mol } CO_2}{1\text{ mol Urea}} = 18.7\text{ mol } CO_2$.
  • Convert moles to grams: $18.7\text{ mol } CO_2 \times 44.01\text{ g/mol} = 823.98\text{ g } CO_2$.
  • The textbook value for this usage is rounded to $823\text{ g } CO_2$.
• Step 2: Calculate the mass remaining
  • Subtract the mass used from the initial amount present.
  • $1142\text{ g (initial)} - 823\text{ g (used)} = 319\text{ g } CO_2$.
• Result for Part C: There are $319\text{ g}$ of carbon dioxide remaining at the end of the reaction.