Problem 74 - Chemistry The Molecular Science

Question

Quicklime, \(\mathrm{CaO},\) is formed when calcium hydroxide is heated. $$\mathrm{Ca}(\mathrm{OH})_{2}(\mathrm{~s}) \longrightarrow \mathrm{CaO}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\ell)$$ The theoretical yield is \(65.5 \mathrm{~g}\) but only \(36.7 \mathrm{~g}\) quicklime is produced. Calculate the percent yield.

Step-by-Step Solution

Answer

The percent yield of quicklime is approximately 56.03%.

Step 1: Understand the Percent Yield Formula

Percent yield is calculated using the formula: \[ \text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100\% \] where \(\text{Actual Yield}\) is the amount of product actually obtained from the reaction, and \(\text{Theoretical Yield}\) is the amount of product expected from the reaction according to stoichiometric calculations.

Step 2: Identify Actual and Theoretical Yields

From the given exercise, the theoretical yield of \(\mathrm{CaO}\) is \(65.5 \mathrm{~g}\), and the actual yield of \(\mathrm{CaO}\) is \(36.7 \mathrm{~g}\). These values will be used in the percent yield formula.

Step 3: Substitute Values into the Percent Yield Formula

Substitute the actual yield and theoretical yield into the percent yield formula: \[ \text{Percent Yield} = \left( \frac{36.7}{65.5} \right) \times 100\% \]

Step 4: Calculate the Percent Yield

First, calculate the fraction: \( \frac{36.7}{65.5} \approx 0.5603 \). Then, multiply by 100 to convert this fraction into a percentage. \[ \text{Percent Yield} = 0.5603 \times 100\% = 56.03\% \]

Step 5: Round the Percent Yield

Round the percent yield to an appropriate number of significant figures. In this case, it can be rounded to two decimal places: \[ \text{Percent Yield} \approx 56.03\% \]

Key Concepts

Theoretical Yield

The theoretical yield is a critical concept in chemistry that represents the maximum amount of product that could be generated from a given chemical reaction, assuming perfect conditions. It is based on the stoichiometric calculations derived from the balanced chemical equation. This yield is calculated using the masses of the reactants and the molar ratios between the reactants and products. When a chemist starts with a specific amount of reactants, they can use stoichiometry to predict how much product should be obtained if the reaction goes to completion with no losses.
In the context of this exercise, the theoretical yield of calcium oxide (\(\mathrm{CaO}\)) is given as \(65.5\ \mathrm{g}\). This value comes from calculating how much \(\mathrm{CaO}\) should be produced based on the amount of calcium hydroxide (\(\mathrm{Ca(OH)}_2\)) used and the balanced chemical equation reflecting the 1:1 molar conversion from \(\mathrm{Ca(OH)}_2\) to \(\mathrm{CaO}\).
It's important to note that the theoretical yield acts as a benchmark, helping chemists determine the efficiency of the actual reaction by comparing it to the actual yield.

Actual Yield

Actual yield refers to the amount of product that is actually obtained from a chemical reaction. Due to various factors like incomplete reactions, side reactions, or losses during the extraction of the product, the actual yield is often less than the theoretical yield. This can occur due to issues with reaction conditions, such as temperature or pressure, or due to experimental errors.
In our exercise, the actual yield of \(\mathrm{CaO}\), the quicklime produced, is \(36.7\ \mathrm{g}\). This is the measurable quantity obtained, as opposed to the ideal amount predicted by calculations. Understanding the difference between actual yield and theoretical yield is crucial for calculating the percent yield. The actual yield is used as the numerator in the percent yield formula, helping to understand how effective the experimental procedure was compared to the theoretical expectations.

Stoichiometry

Stoichiometry is a section of chemistry that involves quantitative relationships in chemical reactions. At its core, stoichiometry helps chemists calculate how much of each reactant is needed and how much product will be formed using balanced chemical equations. These calculations are essential for determining both theoretical and actual yields.
The balanced chemical equation, \(\mathrm{Ca(OH)}_2(\mathrm{s}) \longrightarrow \mathrm{CaO}(\mathrm{s})+\mathrm{H}_2 \mathrm{O}(\ell)\), illustrates the stoichiometric relationship between reactants and products in our example. Here, one mole of calcium hydroxide decomposes to form one mole of calcium oxide and one mole of water. This 1:1 molar ratio is vital for calculating the theoretical yield, as it allows chemists to predict the maximum amount of product that can form from a given amount of reactants.
By mastering stoichiometry, students can efficiently convert between moles and grams and make reliable predictions about the progress and output of chemical reactions. Understanding these relationships lays the foundation for calculating realistic yields and efficiencies in laboratory settings.