Summary of Lecture
In this class, we discussed the concepts of limiting reactants, excess reactants, theoretical yield, and percent yield in chemical reactions. We primarily used two example reactions to explain these concepts: the combustion of propane with oxygen to form carbon dioxide and water, and the combustion of benzene with oxygen to form water and carbon dioxide.
Key Points and Concepts
Balancing Chemical Reactions
- Combustion Reactions:
- Example 1: Propane (C3H8) and Oxygen (O2) to Carbon Dioxide (CO2) and Water (H2O)
- Balancing Steps: Balance C atoms first > H atoms > O atoms last.
- Balanced Equation: C3H8 + 5O2 -> 3CO2 + 4H2O
- Example 2: Benzene (C6H6) and Oxygen to Water and Carbon Dioxide
- Balancing Steps: Done similarly—start with C, then H, then O.
- Balanced Equation: Adjusted to avoid fractional coefficients (doubled all coefficients).
Identifying Limiting and Excess Reactants
- Identifying Limiting Reactant:
- Done by comparing mole ratios from the balanced equation to actual available moles.
- Use ratios ( \frac{\text{given moles}}{\text{coefficient in balanced equation}} ) to find which reactant runs out first (smaller ratio).
- Example: Propane and Oxygen
- Propane ratio = 2/1 = 2
- Oxygen ratio = 8/5 = 1.6
- Oxygen is the limiting reactant as it has a smaller ratio.
Calculating Theoretical Yield
- Derived from the limiting reactant using mole-to-mole conversions from the balanced equation.
- Example: From Oxygen to Carbon Dioxide in propane combustion:
- Molar ratio (O2 to CO2) = 5:3
- If all 8 moles of O2 react, ( \frac{8}{5} \times 3 = 4.8 ) moles of CO2 are the theoretical yield.
Determining Percent Yield
- Formula: ( \text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100% )
- Example: If experimentally 4.5 moles of CO2 were produced:
- Percent Yield = ( \left( \frac{4.5}{4.8} \right) \times 100% \approx 93.75% )
Calculating Excess Reactant Leftover
- Determine how much of the excess reactant (non-limiting) participates in the reaction, subtract it from the original amount.
- Method: Convert the amount of limiting reactant reacted to the amount of excess reactant it would react with—subtract this from the original amount of excess reactant.
- Example: For propane in the first reaction,
- Convert 8 moles of O2 (limiting) to moles of propane reacted using mole ratio.
- Propane reacts: ( \frac{8}{5} = 1.6 ) moles. Starting from 2 moles, 0.4 moles remain.
Practical Examples
- Given practical examples help in understanding these theoretical concepts applied to real chemical reactions, especially for tasks like calculating excess reactants and understanding the discrepancies between theoretical and actual yields due to experimental errors.