Problem 92 - Chemistry Matter and Change

Question

Upon heating, calcium carbonate \(\left(\mathrm{CaCO}_{3}\right)\) decomposes to calcium oxide \((\mathrm{CaO})\) and carbon dioxide \(\left(\mathrm{CO}_{2}\right)\) \begin{equation} \begin{array}{l}{\text { a. Determine the theoretical yield of } \mathrm{CO}_{2} \text { if } 235.0 \mathrm{g} \text { of }} \\\ {\mathrm{CaCO}_{3} \text { is heated. }}\end{array} \end{equation} \begin{equation} \begin{array}{l}{\text { b. What is the percent yield of } \mathrm{CO}_{2} \text { if } 97.5 \mathrm{g} \text { of } \mathrm{CO}_{2} \text { is }} \\\ {\text { collected? }}\end{array} \end{equation}

Step-by-Step Solution

Answer

The theoretical yield of CO2 is 103.34 g; the percent yield is 94.36%.

Step 1: Balanced Chemical Equation

Write the balanced chemical equation for the decomposition of calcium carbonate (\(\text{CaCO}_3\)) to calcium oxide (\(\text{CaO}\)) and carbon dioxide (\(\text{CO}_2\)):\[ \text{CaCO}_3 (s) \rightarrow \text{CaO} (s) + \text{CO}_2 (g) \]This equation indicates that 1 mole of \(\text{CaCO}_3\) produces 1 mole of \(\text{CO}_2\).

Step 2: Molar Mass Calculation

Calculate the molar mass of \(\text{CaCO}_3\):- Calcium (Ca): 40.08 g/mol- Carbon (C): 12.01 g/mol- Oxygen (O): 16.00 g/mol\[ \text{Molar mass of } \text{CaCO}_3 = 40.08 + 12.01 + 3(16.00) = 100.09 \text{ g/mol} \]

Step 3: Moles of CaCO3

Determine the number of moles of \(\text{CaCO}_3\) in 235.0 g:\[ \text{Moles of } \text{CaCO}_3 = \frac{235.0 \text{ g}}{100.09 \text{ g/mol}} = 2.348 \text{ mol} \]

Step 4: Moles of CO2 Produced

Because the reaction produces 1 mole of \(\text{CO}_2\) per mole of \(\text{CaCO}_3\), 2.348 moles of \(\text{CaCO}_3\) will produce:\[ 2.348 \text{ mol of CO}_2 \text{ (theoretical)} \]

Step 5: Mass of Theoretical CO2

Calculate the theoretical mass of \(\text{CO}_2\) using its molar mass:- Molar mass of \(\text{CO}_2\): (12.01 + 2 \times 16.00) g/mol = 44.01 g/mol\[ \text{Mass of } \text{CO}_2 = 2.348 \text{ mol} \times 44.01 \text{ g/mol} = 103.34 \text{ g} \]

Step 6: Percent Yield Calculation

Given 97.5 g of \(\text{CO}_2\) is collected, the percent yield is calculated by:\[ \text{Percent Yield} = \left( \frac{97.5 \text{ g}}{103.34 \text{ g}} \right) \times 100\% = 94.36\% \]

Key Concepts

Balanced Chemical Equation

In chemical reactions, a balanced chemical equation is crucial for understanding reactant-product relationships. It represents the conserved nature of mass and the rearrangement of atoms. For calcium carbonate (CaCO_3) decomposing to calcium oxide (CaO) and carbon dioxide (CO_2), the balanced equation is: \[\text{CaCO}_3 (s) \rightarrow \text{CaO} (s) + \text{CO}_2 (g)\]

• This equation tells us that one molecule of CaCO_3 yields one molecule of CO_2.

This balance ensures the conservation of mass and atoms, which is foundational in chemical calculations. Thus, balancing chemical equations is the first essential step in solving stoichiometric problems.

Molar Mass

Molar mass is the weight of one mole of a substance and is expressed in grams per mole (g/mol). It serves as a bridge between the mass of a substance and the amount of substance (in moles). For our reaction, the molar masses are calculated as follows:

• Calcium (Ca): 40.08 g/mol
• Carbon (C): 12.01 g/mol
• Oxygen (O): 16.00 g/mol

Using these values, you can find the molar mass of CaCO_3: \[ 40.08 + 12.01 + 3 \times 16.00 = 100.09 \text{ g/mol}\]
Understanding molar mass is vital as it allows you to convert between grams and moles, enabling further calculations in reaction stoichiometry.

Percent Yield

Percent yield quantifies the efficiency of a chemical reaction. It compares the amount of product actually obtained from a reaction to the theoretical maximum predicted by stoichiometry.
The formula to calculate percent yield is:
\[\text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100\%\]
For the decomposition of CaCO_3, if 97.5 g of CO_2 is collected and 103.34 g is the theoretical amount generated, the percent yield is:
\[\left( \frac{97.5}{103.34} \right) \times 100\% = 94.36\%\]
A percent yield close to 100% indicates high efficiency, while lower values suggest potential losses or errors during the reaction.

Stoichiometry

Stoichiometry is the study of quantitative relationships between substances involved in chemical reactions. It allows chemists to predict how much reactant is needed and the amount of product formed.
In the decomposition reaction of CaCO_3 , stoichiometry helps us determine how much CO_2 will be generated from a given amount of CaCO_3 :

• Firstly, calculate moles of CaCO_3 using its mass and molar mass.
• Then, use the balanced equation to find the molar ratio, which tells us the moles of CO_2 produced.

This ensures precise calculations that are essential for practical applications in laboratories and industries.