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 correct volume of \(CO_{2}\) produced is 4.48 L.

Step-by-step solution

Understanding the Chemical Equation

The decomposition reaction of calcium carbonate (CaCO_{3}) is given as:\[CaCO_{3}(s) \rightarrow CaO(s) + CO_{2}(g)\]From the equation, 1 mole of \(CaCO_{3}\) produces 1 mole of \(CO_{2}\).

Calculate Molar Mass of \(CaCO_{3}\)

The molar mass of \(CaCO_{3}\) is calculated as follows:- Calcium (Ca) has a molar mass of \(40 \, \text{g/mol}\).- Carbon (C) has a molar mass of \(12 \, \text{g/mol}\).- Oxygen (O) has a molar mass of \(16 \, \text{g/mol}\), and since there are 3 oxygen atoms, it becomes \(48 \, \text{g/mol}\).Thus, the molar mass of \(CaCO_{3}\) is:\[40 + 12 + 48 = 100 \, \text{g/mol}\]

Determine Moles of \(CaCO_{3}\)

We are given 20 g of \(CaCO_{3}\). Using the molar mass, the moles of \(CaCO_{3}\) can be calculated as follows:\[\text{moles of } CaCO_{3} = \frac{20 \, \text{g}}{100 \, \text{g/mol}} = 0.2 \, \text{mol}\]

Calculate Volume of \(CO_{2}\) Produced

At STP (Standard Temperature and Pressure), 1 mole of gas occupies 22.4 L. Since 1 mole of \(CaCO_{3}\) produces 1 mole of \(CO_{2}\), the volume of \(CO_{2}\) from 0.2 moles is:\[\text{Volume of } CO_{2} = 0.2 \, \text{mol} \times 22.4 \, \text{L/mol} = 4.48 \, \text{L}\]

Final Answer Selection

Refer to the provided options:(a) \(2.24 \, \text{L}\)(b) \(4.48 \, \text{L}\)(c) \(44.8 \, \text{L}\)(d) \(48.4 \, \text{L}\)The correct answer based on the calculation is choice (b): \(4.48 \, \text{L}\).

Key concepts

Molar Mass Calculation

Molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). It’s essentially the weight of 6.022 x 10²³ particles of that substance, which is Avogadro’s number. Understanding how to determine the molar mass is crucial for converting between grams and moles during chemical reactions.

When calculating the molar mass, we add the atomic masses of each element within a compound based on the periodic table.

• Calcium (Ca) has an atomic mass of 40 g/mol.
• Carbon (C) has an atomic mass of 12 g/mol.
• Oxygen (O) is 16 g/mol, and in the case of CaCO₃, there are three oxygens contributing 48 g/mol.

This brings the molar mass of calcium carbonate (CaCO₃) to 100 g/mol. This value acts as a conversion factor between the grams of the substance and moles during stoichiometric calculations.

Gas Laws at STP

Gas laws describe the behavior of gases under varying conditions of temperature, pressure, and volume. STP, or Standard Temperature and Pressure, is a common reference point in chemistry, defined as 0°C (273.15 K) and 1 atmosphere (atm) pressure.

Under these conditions, 1 mole of an ideal gas occupies a volume of 22.4 liters. This relationship is derived from the Ideal Gas Law, expressed as:\[ PV = nRT \]where P is pressure, V is volume, n is moles, R is the universal gas constant, and T is temperature in Kelvin.

Using this principle, if you know the moles of a gas under STP conditions, you can easily calculate its volume by multiplying the moles by 22.4 L/mol. For example, in the decomposition of calcium carbonate in our original exercise, 0.2 moles of CO₂ results in a volume of 4.48 L.

Chemical Equation Balancing

Balancing chemical equations is a fundamental skill in chemistry that ensures the law of conservation of mass is obeyed. This law states that matter cannot be created or destroyed, so the amount of each element must be the same on both sides of the chemical equation.

To balance an equation, follow these guidelines:

• Write the correct formulas for all reactants and products.
• Count the number of atoms of each element on both sides.
• Adjust coefficients to balance atoms for each element. Starting with the most complex molecule can help simplify the process.
• Ensure all coefficients are in the smallest possible ratios.

In our exercise with calcium carbonate, the equation was already balanced: \[CaCO_{3}(s) \rightarrow CaO(s) + CO_{2}(g)\]
This indicates that one mole of CaCO₃ decomposes to produce one mole each of CaO and CO₂, making the equation straightforward yet perfectly balanced for our purposes.