Question

From the complete decomposition of \(20 \mathrm{~g} \mathrm{CaCO}_{3}\) at STP the volume of \(\mathrm{CO}_{2}\) obtained is: (a) \(2.24 \mathrm{~L}\) (b) \(4.48 \mathrm{~L}\) (c) \(44.8 \mathrm{~L}\) (d) \(48.4 \mathrm{~L}\)

Short answer

The volume of \(\mathrm{CO}_{2}\) obtained is \(4.48 \mathrm{~L}\).

Step-by-step solution

Write the balanced chemical equation

The decomposition of calcium carbonate \(\mathrm{CaCO}_{3}\) yields calcium oxide \(\mathrm{CaO}\) and carbon dioxide \(\mathrm{CO}_{2}\). The balanced chemical equation is: \[ \mathrm{CaCO}_{3}(s) \rightarrow \mathrm{CaO}(s) + \mathrm{CO}_{2}(g) \] This equation tells us that one mole of \(\mathrm{CaCO}_{3}\) decomposes to produce one mole of \(\mathrm{CO}_{2}\).

Calculate molar masses

Calculate the molar mass of \(\mathrm{CaCO}_{3}\). Molar mass of calcium \(\mathrm{Ca}\) is approximately \(40.08 \mathrm{~g/mol}\), carbon \(\mathrm{C}\) is \(12.01 \mathrm{~g/mol}\), and oxygen \(\mathrm{O}\) is \(16.00 \mathrm{~g/mol}\). Add them: \[ \mathrm{CaCO}_{3} = 40.08 + 12.01 + 3(16.00) = 100.09 \mathrm{~g/mol} \]

Determine moles of \(\mathrm{CaCO}_{3}\)

Find the number of moles of \(\mathrm{CaCO}_{3}\) by dividing the given mass by its molar mass: \[ \text{Moles of } \mathrm{CaCO}_{3} = \frac{20 \mathrm{~g}}{100.09 \mathrm{~g/mol}} \approx 0.2 \text{ moles} \]

Relate moles of \(\mathrm{CaCO}_{3}\) to \(\mathrm{CO}_{2}\)

From the balanced equation, 1 mole of \(\mathrm{CaCO}_{3}\) produces 1 mole of \(\mathrm{CO}_{2}\). Therefore, 0.2 moles of \(\mathrm{CaCO}_{3}\) will produce 0.2 moles of \(\mathrm{CO}_{2}\).

Calculate volume of \(\mathrm{CO}_{2}\) at STP

At standard temperature and pressure (STP), 1 mole of a gas occupies \(22.4 \mathrm{~L}\). Therefore, the volume of \(\mathrm{CO}_{2}\) produced is: \[ \text{Volume of } \mathrm{CO}_{2} = 0.2 \text{ moles} \times 22.4 \mathrm{~L/mol} \approx 4.48 \mathrm{~L} \]

Key concepts

Chemical Reactions

Chemical reactions involve the rearrangement of atoms to form new substances. In the context of stoichiometry, it’s essential to understand how reactants transform into products. In the exercise above, calcium carbonate (\(\text{CaCO}_3\)) undergoes decomposition. This is a specific type of reaction where a compound breaks down into two or more simpler substances.

The balanced chemical equation given is:\[ \text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g) \]This indicates that one mole of calcium carbonate decomposes to produce one mole of calcium oxide and one mole of carbon dioxide. Writing the balanced equation is crucial as it tells us the exact ratio in which reactants react and products form. It’s the heart of stoichiometric calculations.

Molar Mass Calculation

To proceed with chemical calculations, we need to know the molar masses of the substances involved. A molar mass is essentially the mass of one mole of a substance. To find the molar mass of calcium carbonate (\(\text{CaCO}_3\)), we add the atomic masses of its component elements:

• Calcium (\(\text{Ca}\)): approximately 40.08 g/mol
• Carbon (\(\text{C}\)): 12.01 g/mol
• Oxygen (\(\text{O}\)), which is found three times in the compound: 3 \(\times\) 16.00 g/mol = 48.00 g/mol

Adding these gives:\[\text{Molar mass of } \text{CaCO}_3 = 40.08 + 12.01 + 48.00 = 100.09 \text{ g/mol}\] Calculating molar mass allows us to convert a substance's mass to moles, facilitating further stoichiometric calculations.

Gas Laws

Gas laws help us understand how gases behave under different conditions. A fundamental concept in stoichiometry involving gases is the use of molar volume at standard conditions. The molar volume is the volume occupied by one mole of a gas.

In this exercise, knowing the mole-volume relationship is crucial. For example, 1 mole of a gas at STP will occupy 22.4 liters. Thus, when given moles, we can easily calculate the volume of carbon dioxide (\text{\(\text{CO}_2\)}) produced using the equation:\[\text{Volume of } \text{CO}_2 = \text{Moles of } \text{CO}_2 \times 22.4 \text{ L/mol}\] Using this relationship, 0.2 moles of \(\text{CO}_2\) will occupy 4.48 liters at STP.

Standard Temperature and Pressure (STP)

Standard temperature and pressure (STP) is a reference point used in chemistry to define one mole of a gas. At STP, conditions are designated as a temperature of 273.15 K (0°C) and a pressure of 1 atm. These conditions provide a standardized way to measure gas volumes.

The concept is particularly useful because it simplifies calculations with gases. At STP, gases have a predictable behavior, where one mole occupies 22.4 liters. Understanding STP is key to solving problems like the one given, as it allows us to determine the volume of gases produced or consumed in a reaction based on the number of moles involved.

By using STP conditions, we can easily convert between moles and volumes, as seen in determining that 0.2 moles of \(\text{CO}_2\) translate to 4.48 liters.