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 \(\mathrm{CO}_2\) released is \(1.12 \mathrm{~L}\).

Step 1: Understand the Decomposition Reaction

When barium carbonate (\(\mathrm{BaCO}_3\)) is heated, it decomposes to form barium oxide (\(\mathrm{BaO}\)) and carbon dioxide (\(\mathrm{CO}_2\)) gas. The chemical reaction is: \( \mathrm{BaCO}_3 \rightarrow \mathrm{BaO} + \mathrm{CO}_2 \). We need to find out how much \(\mathrm{CO}_2\) gas is released at STP (Standard Temperature and Pressure) from \(9.85\, \mathrm{g}\) of barium carbonate.

Step 2: Calculate Molar Mass of BaCO3

The molar masses of the elements are: \(\mathrm{Ba} = 137\, \mathrm{g/mol}\), \(\mathrm{C} = 12\, \mathrm{g/mol}\), and \(\mathrm{O} = 16\, \mathrm{g/mol}\). Therefore, the molar mass of \(\mathrm{BaCO}_3\) is:\[ \text{Molar mass of } \mathrm{BaCO}_3 = 137 + 12 + (16 \times 3) = 197 \, \text{g/mol} \]

Step 3: Determine Moles of BaCO3

To find the moles of \(\mathrm{BaCO}_3\) in \(9.85\, \mathrm{g}\), use the formula:\[ \text{Moles of } \mathrm{BaCO}_3 = \frac{\text{Mass}}{\text{Molar mass}} = \frac{9.85\, \mathrm{g}}{197\, \mathrm{g/mol}} \approx 0.05\, \text{mol} \]

Step 4: Calculate Moles of CO2 Produced

From the balanced equation, each mole of \(\mathrm{BaCO}_3\) produces one mole of \(\mathrm{CO}_2\). Thus, \(0.05\, \text{mol}\) of \(\mathrm{BaCO}_3\) will produce \(0.05\, \text{mol}\) of \(\mathrm{CO}_2\).

Step 5: Determine Volume of CO2 at STP

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

Step 6: Compare with Options

The calculated volume of \(\mathrm{CO}_2\) gas at STP is \(1.12\, \mathrm{L}\), which matches option (a).

Key Concepts

Molar Mass Calculation

When dealing with chemical reactions, knowing the molar mass of a compound is essential to determine how many moles are present in a given mass. The molar mass is simply the sum of the atomic masses of all the atoms in a molecule. For instance, in barium carbonate (\( \mathrm{BaCO}_3 \)), we calculate the molar mass by adding:

• One barium (\( \mathrm{Ba} \)), which is \( 137 \, \mathrm{g/mol} \)
• One carbon (\( \mathrm{C} \)), which is \( 12 \, \mathrm{g/mol} \)
• Three oxygen atoms (\( 3 \times 16 = 48 \, \mathrm{g/mol} \))

Ultimately, the molar mass of \( \mathrm{BaCO}_3 \) is computed to be \( 197 \, \mathrm{g/mol} \). Once we know the molar mass, it's straightforward to convert a known mass of substance into moles, using the formula:\[\text{Moles} = \frac{\text{Mass of substance}}{\text{Molar mass}}\]This calculation helps us determine how much of a substance is present in a chemical reaction, which is pivotal for further analysis.

Gas Laws at STP

Standard Temperature and Pressure (STP) is a set reference point in science to denote the condition of 0°C (273.15 K) for temperature and 1 atm pressure. This standard allows scientists to make consistent calculations about gases. A crucial concept at STP is that one mole of any ideal gas occupies a volume of \( 22.4 \, \mathrm{L} \).So whenever you're dealing with reactions involving gases, this information allows you to convert from moles to liters and vice versa easily under STP conditions. For example, when \( \mathrm{BaCO}_3 \) decomposes and releases \( \mathrm{CO}_2 \) at STP, each mole of \( \mathrm{CO}_2 \) takes up \( 22.4 \, \mathrm{L} \). If you know the number of moles released, the volume can be calculated as follows:\[\text{Volume of gas at STP} = \text{Number of moles} \times 22.4 \, \mathrm{L/mol}\]This simple relationship makes predictions and calculations in stoichiometry straightforward.

Chemical Reactions

Chemical reactions are processes where substances, known as reactants, transform into different substances, called products. In the case of barium carbonate (\( \mathrm{BaCO}_3 \)), the decomposition reaction produces barium oxide (\( \mathrm{BaO} \)) and carbon dioxide (\( \mathrm{CO}_2 \)). This can be represented by the chemical equation:\[\mathrm{BaCO}_3 \rightarrow \mathrm{BaO} + \mathrm{CO}_2\]Each part of this equation provides useful information:

• The reactants and products are clearly defined.
• It indicates a 1:1 molar ratio between reactant and products.

Understanding chemical reactions involves recognizing patterns in the way substances react. The law of conservation of mass governs these reactions, meaning the total mass of the reactants equals the total mass of the products. This concept ensures that equations are balanced.In this context, calculating the amount of gas released, or products formed, hinges on knowing the initial conditions and using the stoichiometry derived from these balanced equations.