Problem 43 - Chemistry Matter and Change

Question

When solid calcium carbonate \(\left(CaCO_{3}\right)\) is heated, it decomposes to form solid calcium oxide (CaO) and carbon dioxide gas \(\left(CO_{2}\right) .\) How many liters of carbon dioxide will be produced at STP if 2.38 \(kg\) of calcium carbonate reacts completely?

Step-by-Step Solution

Answer

505.12 liters of carbon dioxide will be produced at STP if 2.38 kg of calcium carbonate reacts completely.

Step 1: Write the balanced chemical equation

The balanced chemical equation for the decomposition of calcium carbonate is: \[CaCO_3(s) \rightarrow CaO(s) + CO_2(g)\]

Step 2: Convert the mass of calcium carbonate into moles

Now, we need to convert the given mass of calcium carbonate (2.38 kg) into moles. To do that, we'll use the molar mass of calcium carbonate, which is: Molar mass of \(CaCO_3\) = 40.08 (Ca) + 12.01 (C) + 3×16.00 (O) = 100.09 g/mol First, convert the mass from kg to grams: Mass of \(CaCO_3 = 2.38 kg × \frac{1000 g}{1 kg} = 2380 g\) Now, convert the mass to moles: Moles of \(CaCO_3 = \frac{2380 g}{100.09 g/mol} = 23.77 mol\)

Step 3: Use stoichiometry to find the moles of carbon dioxide produced

As per the balanced chemical equation, one mole of calcium carbonate (CaCO3) decomposes to produce one mole of carbon dioxide gas (CO2). So, the moles of carbon dioxide produced are equal to the moles of calcium carbonate reacted: Moles of \(CO_2 = 23.77 mol\)

Step 4: Convert the moles of carbon dioxide into liters at STP

In order to convert the moles of carbon dioxide into liters at STP, we'll use the ideal gas law equation: \[PV = nRT\] Where: - P is the pressure (in atm) - V is the volume (in liters) - n is the number of moles - R is the ideal gas constant (\(0.0821 \frac{L⋅atm}{mol⋅K}\)) - T is the temperature (in Kelvin) At STP, the pressure (P) is 1 atm and the temperature (T) is 273.15 K. So, rearrange the equation to solve for the volume (V): \[V = \frac{nRT}{P}\] Now, substitute the values and solve: V = \(\frac{23.77 mol × 0.0821 \frac{L⋅atm}{mol⋅K} × 273.15 K}{1 atm} = 505.12 L\) Therefore, 505.12 liters of carbon dioxide will be produced at STP if 2.38 kg of calcium carbonate reacts completely.

Key Concepts

Chemical Reactions

Chemical reactions are processes where substances, called reactants, transform into different substances, known as products. In the case of calcium carbonate (\(CaCO_3\)), when it is heated, it undergoes a decomposition reaction. This specific reaction breaks down a single compound into simpler products.

In our example, calcium carbonate decomposes into calcium oxide \(CaO\) and carbon dioxide \(CO_2\). The balanced chemical equation for this reaction is:\[CaCO_3(s) \rightarrow CaO(s) + CO_2(g)\]

• This equation shows the starting materials (reactants) on the left and the end products on the right.
• A balanced equation indicates that the number of each type of atom is the same on both sides of the equation. This reflects the Law of Conservation of Mass.

Understanding these basics helps us predict the products of a reaction and the quantities involved.

Mole Concept

The mole is a fundamental concept in chemistry that connects the mass of substances to the number of particles they contain. This is essential for quantifying chemicals and reactions.

When working with chemical equations, we use moles to understand how substances react at the molecular level.

For the decomposition of calcium carbonate, the mass should be converted from kilograms to grams before finding the number of moles, since molar masses are given in grams per mole. The conversion steps were:

• Convert kilograms to grams by multiplying: \(2.38 \, \text{kg} \times 1000 = 2380 \, \text{g}\).
• Find moles by dividing the mass by molar mass: \( \frac{2380 \, \text{g}}{100.09 \, \text{g/mol}} = 23.77 \, \text{mol}\) of \(CaCO_3\).

Moles link the macroscopic world (quantities we measure) to the microscopic world (atoms and molecules), allowing us to calculate proportions in reactions accurately.

Ideal Gas Law

The ideal gas law is a crucial equation in chemistry for understanding the behavior of gases. It relates the pressure, volume, temperature, and number of moles of a gas:\[PV = nRT\]

• \(P\) is the pressure of the gas.
• \(V\) is the volume.
• \(n\) is the number of moles.
• \(R\) is the ideal gas constant \(0.0821 \frac{L \cdot atm}{mol \cdot K}\)
• \(T\) is the temperature in Kelvin.

At standard temperature and pressure (STP: 1 atm and 273.15 K), calculations become relatively straightforward.
Using the number of moles of \(CO_2\) from our reaction (23.77 mol), the volume can be calculated by rearranging the ideal gas law to:\[V = \frac{nRT}{P}\]Substituting the known values, we find:\[V = \frac{23.77 \times 0.0821 \times 273.15}{1} = 505.12 \, \text{L}\]This volume of carbon dioxide shows how the ideal gas law bridges chemical reactions and real-world measurements by using gas properties.