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 at STP is \(1.12\, \text{L}\).

Step 1: Determine Molar Mass of BaCO3

First, calculate the molar mass of \(\text{BaCO}_3\). The atomic masses are: \(\text{Ba} = 137\, \text{g/mol}\), \(\text{C} = 12\, \text{g/mol}\), \(\text{O} = 16\, \text{g/mol}\). \(\text{BaCO}_3\) has one barium atom, one carbon atom, and three oxygen atoms. Thus, the molar mass of \(\text{BaCO}_3\) is calculated as follows:\[\text{Molar mass of BaCO}_3 = 137 + 12 + (3 \times 16) = 197\, \text{g/mol}.\]

Step 2: Calculate Moles of BaCO3

Using the given mass of \(\text{BaCO}_3\), we calculate the number of moles using the formula:\[\text{moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{9.85}{197}.\]

Step 3: Determine Moles of CO2 Produced

The decomposition of \(\text{BaCO}_3\) is given by the reaction:\[\text{BaCO}_3 \rightarrow \text{BaO} + \text{CO}_2.\]This implies that 1 mole of \(\text{BaCO}_3\) will produce 1 mole of \(\text{CO}_2\). Thus, the moles of \(\text{CO}_2\) are equal to the moles of \(\text{BaCO}_3\), which is \(\frac{9.85}{197}\).

Step 4: Calculate Volume of CO2 at STP

At standard temperature and pressure (STP), 1 mole of any gas occupies \(22.4\, \text{L}\). Therefore, the volume of \(\text{CO}_2\) produced is:\[\text{Volume of } \text{CO}_2 = \text{moles of } \text{CO}_2 \times 22.4\, \text{L/mol} = \left(\frac{9.85}{197}\right) \times 22.4 \approx 1.12\, \text{L}.\]

Key Concepts

Molar Mass Calculation

Calculating the molar mass of a compound is a key step in stoichiometry, which helps us to relate mass with moles. The molar mass is simply the sum of the atomic masses of all the atoms in a molecule expressed in grams per mole (g/mol). For instance, to calculate the molar mass of barium carbonate, \( \text{BaCO}_3 \), we need the atomic masses of its constituent elements as listed:

• Barium (Ba): 137 g/mol
• Carbon (C): 12 g/mol
• Oxygen (O): 16 g/mol

Since there's one Ba, one C, and three O atoms in \( \text{BaCO}_3 \), the molar mass calculation is as follows:
137 + 12 + (3 \times 16) = 197 \, \text{g/mol}.
Understanding molar mass enables you to convert between grams of a substance and moles, which is crucial in solving stoichiometry problems efficiently.

Gas Volume at STP

In chemistry, Standard Temperature and Pressure (STP) is a common reference point used to measure gas volumes. At STP, one mole of any ideal gas occupies a volume of 22.4 liters. This uniformity simplifies calculations when dealing with gaseous substances.
For example, when calculating the volume of carbon dioxide \( \text{CO}_2 \) produced from a reaction, knowing that each mole of \( \text{CO}_2 \) at STP will occupy 22.4 L is extremely useful.
You can determine the volume of gas produced by multiplying the number of moles of gas by 22.4 L/mol. Practically speaking, it means if we have \( \text{moles} \) of \( \text{CO}_2 = \frac{9.85}{197} \), then:\[\text{Volume of } \text{CO}_2 = \left(\frac{9.85}{197}\right) \times 22.4 \, \text{L/mol} \approx 1.12 \, \text{L}.\]This calculation demonstrates how powerful simple conversions can be in predicting results in chemical reactions.

Chemical Reactions

Chemical reactions describe the process by which substances interact to form new products. In the case of barium carbonate, \( \text{BaCO}_3 \), it decomposes upon heating to form barium oxide, \( \text{BaO} \), and carbon dioxide, \( \text{CO}_2 \). The balanced chemical equation for this decomposition is:
\[\text{BaCO}_3 \rightarrow \text{BaO} + \text{CO}_2\]
Here, the equation states that one mole of \( \text{BaCO}_3 \) yields one mole each of \( \text{BaO} \) and \( \text{CO}_2 \). This relationship helps us in calculating the quantities of reactants and products involved. Since the decomposition releases \( \text{CO}_2 \), knowing about this fixed stoichiometric relationship allows us to predict the amount of \( \text{CO}_2 \) generated based on the initial quantity of \( \text{BaCO}_3 \). It becomes insightful to recognize that stoichiometry not only deals with the quantities of matter but also provides an understanding of the conservation of mass and atoms in a chemical reaction.