Summary of "3.4 Limiting Reactant Problems | General Chemistry"

This lesson focuses on understanding and solving Limiting Reactant (limiting reagent) problems in Stoichiometry, a key topic in general chemistry. The instructor, Chad, guides students through the concepts, calculations, and methodologies needed to confidently approach these problems, using clear examples and step-by-step dimensional analysis.

Main Ideas and Concepts

Stoichiometry and Limiting Reactants:
- Stoichiometry involves mole-to-mole ratios derived from balanced chemical equations.
- Balanced chemical equations provide the ratio of reactants and products but do not specify the actual amounts present.
- The Limiting Reactant is the reactant that runs out first, limiting the amount of product formed.
- The reactant in excess is leftover after the reaction stops.
Mole Ratios:
- Mole ratios come from the coefficients in balanced chemical reactions.
- These ratios allow conversion between moles of different substances in a reaction.
- Example: For Ammonia Synthesis, \( N_2 + 3H_2 \rightarrow 2NH_3 \), the mole ratios are:
  - \(N_2 : H_2 = 1 : 3\)
  - \(N_2 : NH_3 = 1 : 2\)
  - \(H_2 : NH_3 = 3 : 2\)
Calculations with Moles and Grams:
- When given grams, convert to moles using Molar Mass before applying mole ratios.
- The Molar Mass connects grams and moles (e.g., \(N_2\) = 28 g/mol, \(H_2\) = 2 g/mol).
- The process often involves:
  1. Convert grams to moles.
  2. Use mole ratios to find moles of another substance.
  3. Convert moles back to grams if required.
Limiting Reactant Problem Approach:
- Calculate the maximum product amount possible from each reactant individually.
- The smaller product amount indicates the Limiting Reactant.
- The other reactant is in excess.
- Calculate leftover excess reactant by subtracting used amount from initial amount.
Theoretical Yield:
- The maximum amount of product possible based on Limiting Reactant.
- Usually expressed in moles or grams.
- Real reactions often yield less due to side reactions and incomplete reactions.
Percent Yield:
- \[ \text{Percent Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100 \]
- Actual yield is the amount of product actually obtained.
- Percent Yield typically less than 100%; greater than 100% usually indicates impurities (e.g., wet product).

Detailed Methodology / Step-by-Step Instructions

Identify the balanced chemical equation.
- Example: \( N_2 + 3H_2 \rightarrow 2NH_3 \)
Convert all given masses to moles.
- Use molar masses (from periodic table).
- \( \text{moles} = \frac{\text{grams}}{\text{Molar Mass}} \)
Calculate the maximum amount of product from each reactant.
- Use mole ratios from the balanced equation.
- For each reactant:
  - \( \text{moles of product} = \text{moles of reactant} \times \frac{\text{coefficient of product}}{\text{coefficient of reactant}} \)
Determine the Limiting Reactant.
- The reactant that produces the least amount of product is limiting.
- The other reactant is in excess.
Calculate Theoretical Yield.
- The maximum amount of product formed based on Limiting Reactant.
Calculate leftover excess reactant.
- Find moles of excess reactant used to produce Theoretical Yield.
- Subtract used moles from initial moles.
- Convert leftover moles to grams if needed.
Calculate Percent Yield (if actual yield is given).
- \( \text{Percent Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100 \)

Example Problems Covered

Ammonia Synthesis reaction:
- Given moles or grams of \(N_2\) and \(H_2\), find:
  - How many moles of \(H_2\) are needed to react with given \(N_2\).
  - Maximum moles of \(NH_3\) produced.
  - Limiting Reactant when given grams of both reactants.
  - Amount of excess reactant left after reaction.
  - Percent Yield given