Chemistry - Yield of a Reaction in Chemical Equations (24 of 38) : Example By Michel van Biezen

Summary:

1. Introduction to Theoretical Yields and Percent Yields:

   • Topic: Theoretical yields and percent yields in a chemical reaction.
   • Example: Illustrates the chemical reaction involving iron oxide and carbon, commonly used in steel production.

2. Overview of the Reaction:

   • Reaction: Iron oxide reacts with carbon under heat to produce iron and carbon monoxide gas.
   • Context: A step in the steel production process, extracting iron from iron ore (iron iii oxide).

3. Quantities in the Reaction:

   • Initial Quantities: 30,000 kilograms of iron oxide ore and 100,000 kilograms of carbon (coke) are used.
   • Objective: Maximize iron extraction by ensuring ample carbon availability.

4. Theoretical Yield Calculation:

   • Approach: Calculate the theoretical yield by understanding the molar masses involved.
   • Molar Mass: Molar mass of iron is 55.85 grams, and iron iii oxide has a molar mass of 159.7 grams per mole.
   • Calculation: Theoretical yield (${�}_{\text{theoretical}}$) is calculated using the formula: ${�}_{\text{theoretical}}=\left(\frac{\text{mass of iron}}{\text{molar mass of iron iii oxide}}\right)\times \text{amount started with}$
   • Result: Theoretical yield is found to be 20,983 kilograms of iron if all the iron is extracted.

5. Percent Yield Calculation:

   • Introduction: Define percent yield as the ratio of actual yield to theoretical yield, multiplied by 100%.
   • Formula: $\text{Percent Yield}=\left(\frac{\text{Actual Yield}}{\text{Theoretical Yield}}\right)\times 100$
   • Given: Actual yield is 15,000 kilograms.
   • Calculation: $\text{Percent Yield}=\left(\frac{15,000}{20,983}\right)\times 100$
   • Result: Percent yield is calculated to be 71.5%.

6. Conclusion:

   • Interpretation: The steel production process, in this example, achieves a 71.5% yield in extracting iron from the provided iron oxide ore.
   • Consideration: Theoretical yield provides an ideal value, while actual yield determines the practical efficiency of the process.

7. Summary and Application:

   • Recap: Demonstrates how to calculate theoretical yield and percent yield in a chemical reaction.
   • Application: Applicable in various chemical processes to assess the efficiency and yield of desired products.

This summary captures the key steps and calculations involved in understanding theoretical yields and percent yields in the context of a chemical reaction.