Problem 93 - General Chemistry

Question

When calcium carbonate, \(\mathrm{CaCO}_{3}\) (the major constituent of limestone and seashells), is heated, it decomposes to calcium oxide (quicklime). $$ \mathrm{CaCO}_{3}(s) \longrightarrow \mathrm{CaO}(s)+\mathrm{CO}_{2}(g) ; \Delta H=177.9 \mathrm{~kJ} $$ How much heat is required to decompose \(21.3 \mathrm{~g}\) of calcium carbonate?

Step-by-Step Solution

Answer

Approximately 37.9 kJ of heat is required.

Step 1: Determine the Molar Mass of Calcium Carbonate

Calcium carbonate has the formula \( \mathrm{CaCO}_{3} \). To find its molar mass, sum the atomic masses of its constituent elements: 1 calcium (Ca) atom, 1 carbon (C) atom, and 3 oxygen (O) atoms.- Atomic mass of Ca = 40.08 \( \mathrm{g/mol} \) - Atomic mass of C = 12.01 \( \mathrm{g/mol} \) - Atomic mass of O = 16.00 \( \mathrm{g/mol} \) Thus, the molar mass of \( \mathrm{CaCO}_{3} = 40.08 + 12.01 + (3 \times 16.00) = 100.09 \; \mathrm{g/mol} \)

Step 2: Calculate the Moles of Calcium Carbonate

Using the formula to calculate moles, \( \text{moles} = \frac{\text{mass}}{\text{molar mass}} \), we can find the number of moles of \( \mathrm{CaCO_{3}} \) in 21.3 grams:\[ \text{moles of } \mathrm{CaCO_{3}} = \frac{21.3 \; \mathrm{g}}{100.09 \; \mathrm{g/mol}} \approx 0.213 \; \mathrm{mol} \]

Step 3: Calculate the Heat Required for the Reaction

The balanced chemical equation indicates that the decomposition of 1 mole of \( \mathrm{CaCO_{3}} \) requires \( 177.9 \; \mathrm{kJ} \) of heat. Since we have already calculated the moles of \( \mathrm{CaCO_{3}} \) as 0.213 mol, we use the relation:\[ \text{Heat required} = 0.213 \; \mathrm{mol} \times 177.9 \; \mathrm{kJ/mol} \approx 37.9 \; \mathrm{kJ} \]

Key Concepts

Molar Mass Calculation

A molar mass is like a compound's fingerprint. It tells you the mass of one mole of that substance in grams. Calculating the molar mass is straightforward, but it's crucial for many chemical calculations.
To find the molar mass of calcium carbonate, \( \mathrm{CaCO}_{3} \), we add up the atomic masses of each component. Here's what you do:

• Find the atomic mass of calcium (Ca). It's 40.08 grams per mole.
• Add the mass of carbon (C), which is 12.01 grams per mole.
• Add three times the mass of oxygen (O), since there are three oxygen atoms. Each oxygen is 16.00 grams per mole.

Add these masses together: 40.08 + 12.01 + (3 × 16.00) to get a total of 100.09 grams per mole. This is the molar mass of calcium carbonate, and it helps convert grams of the substance into moles.

Enthalpy Change

Enthalpy change is an essential concept in thermodynamics and chemistry. But what does it mean in the context of a reaction?The term "enthalpy change" refers to the heat absorbed or released during a chemical reaction at constant pressure. In our example, the decomposition reaction of calcium carbonate has an enthalpy change or \( \Delta H \) of \( 177.9 \; \mathrm{kJ} \). This means each mole of \( \mathrm{CaCO}_{3} \) that decomposes absorbs 177.9 kilojoules of heat.
If the enthalpy change is positive, the reaction is endothermic (absorbs heat). This is the case here.

• \( \Delta H > 0 \): Endothermic, heat is absorbed.
• \( \Delta H < 0 \): Exothermic, heat is released.

Understanding enthalpy change helps predict how much heat energy is needed or produced in a reaction.

Chemical Reactions

Chemical reactions describe how substances interact to create new products. In our exercise, calcium carbonate decomposes into calcium oxide and carbon dioxide. Let's break down the specifics:
The chemical equation \( \mathrm{CaCO}_{3}(s) \rightarrow \mathrm{CaO}(s)+\mathrm{CO}_{2}(g) \) tells us exactly what's happening.

• \( \mathrm{CaCO}_{3} \) (solid) is breaking down.
• It produces two new substances: \( \mathrm{CaO} \) (another solid, known as quicklime) and \( \mathrm{CO}_{2} \) (a gas).
• The reaction is all about bonds breaking and forming.

For students, understanding chemical reactions means paying attention to changes in physical states and energy, as well as noting products and reactants.

Stoichiometry

Stoichiometry is like a recipe for chemical reactions. It tells us the "ingredients" and their quantities needed for chemical transformations. When calculating how much heat is needed to decompose a given amount of calcium carbonate, this principle is crucial. Based on the chemical equation, stoichiometry informs us that:

• 1 mole of \( \mathrm{CaCO}_{3} \) provides 1 mole of \( \mathrm{CaO} \) and 1 mole of \( \mathrm{CO}_{2} \).
• Per mole of \( \mathrm{CaCO}_{3} \), 177.9 kJ of heat is needed for decomposition.

Now, if you have 21.3 grams of calcium carbonate, first convert this to moles using its molar mass (100.09 g/mol).
Once you know the moles, multiply by the enthalpy change per mole to find the heat required. In this example, 0.213 moles of \( \mathrm{CaCO}_{3} \) require approximately 37.9 kJ of heat, as computed earlier.