Problem 127 - A Complete Resource Book in Chemistry for JEE Main

Question

Assuming fully decomposed, the volume of \(\mathrm{CO}_{2}\) released at STP on heating \(9.85 \mathrm{~g}\) of \(\mathrm{BaCO}_{3}\) (atomic mass, \(\mathrm{Ba}=137\) ) will be (a) \(1.12 \mathrm{~L}\) (b) \(4.84 \mathrm{~L}\) (c) \(2.12 \mathrm{~L}\) (d) \(2.06 \mathrm{~L}\)

Step-by-Step Solution

Answer

The volume of \(\text{CO}_2\) released is \(1.12 \, \text{L}\).

Step 1: Write the chemical equation

The decomposition of barium carbonate, \(\text{BaCO}_3\), can be represented by the equation: \[ \text{BaCO}_3 (s) \rightarrow \text{BaO} (s) + \text{CO}_2 (g) \] This equation shows that 1 mole of \(\text{BaCO}_3\) decomposes to produce 1 mole of \(\text{CO}_2\).

Step 2: Calculate the molar mass of BaCO3

The molar mass of \(\text{BaCO}_3\) is calculated as follows: \[ \text{Molar mass of } \text{BaCO}_3 = 137 + 12 + (3 \times 16) = 137 + 12 + 48 = 197 \, \text{g/mol} \] Thus, the molar mass of \(\text{BaCO}_3\) is \(197 \, \text{g/mol}\).

Step 3: Calculate moles of BaCO3

To find the moles of \(\text{BaCO}_3\) in \(9.85 \, \text{g}\), use the formula: \[ \text{moles of BaCO}_3 = \frac{\text{mass}}{\text{molar mass}} = \frac{9.85}{197} \approx 0.05 \, \text{moles} \] Thus, \(0.05\) moles of \(\text{BaCO}_3\) are present.

Step 4: Calculate moles of CO2 produced

According to the chemical equation, the decomposition of 1 mole of \(\text{BaCO}_3\) produces 1 mole of \(\text{CO}_2\). Therefore, \(0.05\) moles of \(\text{BaCO}_3\) will produce \(0.05\) moles of \(\text{CO}_2\).

Step 5: Calculate volume of CO2 at STP

At standard temperature and pressure (STP), 1 mole of any gas occupies \(22.4 \, \text{L}\). Thus, the volume of \(\text{CO}_2\) produced is: \[ \text{Volume of CO}_2 = 0.05 \, \text{moles} \times 22.4 \, \text{L/mol} = 1.12 \, \text{L} \]

Key Concepts

Chemical Reactions

Chemical reactions describe processes where substances, known as reactants, are transformed into different substances called products. In the case of \(\text{BaCO}_3\), we are looking at a decomposition reaction, which is a type of chemical reaction where a single compound breaks down into two or more elements or new compounds. This is an important aspect of stoichiometry, which is the calculation of reactants and products in chemical reactions.

Here is what happens when \(\text{BaCO}_3\) decomposes:

• The solid \(\text{BaCO}_3\) breaks down into barium oxide, \(\text{BaO}\), a solid, and carbon dioxide, \(\text{CO}_2\), a gas.
• This means that for every mole of \(\text{BaCO}_3\) that decomposes, one mole of \(\text{CO}_2\) is produced.

Understanding the stoichiometry of the reaction is crucial, as it tells us how the reactants relate to the products, which is essential in calculating the volume of gas released in this reaction.

Gas Laws

Gas laws are critical for understanding how gases behave under different conditions. At Standard Temperature and Pressure (STP) — defined as 0°C and 1 atmosphere pressure — one mole of any ideal gas occupies 22.4 liters. This concept comes into play when we calculate the volume of gas released in a reaction.

In our example:

• We apply this principle to determine the amount of \(\text{CO}_2\) produced. Knowing the volume occupied by a mole of gas at STP allows us to directly calculate the volume of gas produced when given the number of moles.
• This means that because 0.05 moles of \(\text{CO}_2\) are produced, the volume of \(\text{CO}_2\) can be calculated using the formula: \[ \text{Volume of } \text{CO}_2 = 0.05 \times 22.4 \text{ L/mol} = 1.12 \text{ L} \]

Practical applications of these gas laws help chemists and engineers predict how gases will behave in different environments, which is crucial in fields such as chemical engineering and atmospheric science.

Molar Mass Calculation

Molar mass is the mass of one mole of a given chemical element or compound, often expressed in grams per mole (g/mol). It is a fundamental concept used to convert between the mass of a substance and the amount in moles, a basic unit of chemical quantity. Calculating the molar mass of a compound involves adding up the atomic masses of the elements that make it up.

For barium carbonate (\text{BaCO}_3):

• Start with the atomic masses: Barium (\text{Ba}) is 137, Carbon (\text{C}) is 12, and Oxygen (\text{O}) is 16.
• Since \(\text{BaCO}_3\) comprises one barium atom, one carbon atom, and three oxygen atoms, the molar mass calculation is: \[ 137 + 12 + (3 \times 16) = 197 \, \text{g/mol} \]
• This value is crucial for stoichiometry calculations, allowing us to determine how much of each reactant is needed to produce a certain amount of product.

Molar mass serves as a bridge between the atomic scale and the macroscopic measurements we use in the laboratory.