Problem 94 - A Complete Resource Book in Chemistry for JEE Main

Question

The mass of \(\mathrm{BaCO}_{3}\) formed where excess of \(\mathrm{CO}_{2}\) is passed through a solution having \(0.205\) mole of Ba \((\mathrm{OH})_{2}\) is? (a) \(40.5 \mathrm{gm}\) (b) \(20.25 \mathrm{gm}\) (c) \(81 \mathrm{gm}\) (d) \(4.05 \mathrm{gm}\)

Step-by-Step Solution

Answer

(a) 40.5 gm

Step 1: Identify the Chemical Reaction

The chemical reaction between barium hydroxide and carbon dioxide to form barium carbonate and water is: \[ \mathrm{Ba(OH)_2 + CO_2 \rightarrow BaCO_3 + H_2O} \] We see that one mole of barium hydroxide reacts with one mole of carbon dioxide to produce one mole of barium carbonate.

Step 2: Moles of Barium Hydroxide

You are given 0.205 moles of \( \mathrm{Ba(OH)_2} \). Since \( \mathrm{CO_2} \) is in excess, all the barium hydroxide will react to form barium carbonate. Thus, 0.205 moles of \( \mathrm{BaCO_3} \) will be produced.

Step 3: Calculate Molar Mass of Barium Carbonate

The molar mass of \( \mathrm{BaCO_3} \) is calculated as follows: Barium (Ba) = 137 g/mol, Carbon (C) = 12 g/mol, and Oxygen (O) = 16 g/mol. Therefore, the molar mass of \( \mathrm{BaCO_3} \) is \( 137 + 12 + (3 \times 16) = 197 \) g/mol.

Step 4: Mass of Barium Carbonate Formed

To find the mass of \( \mathrm{BaCO_3} \) formed, multiply the moles of \( \mathrm{BaCO_3} \) by its molar mass: \[ \text{Mass} = 0.205 \text{ moles} \times 197 \text{ g/mol} = 40.385 \text{ grams} \]

Step 5: Select the Correct Answer

The closest option to our calculated mass (40.385 grams) is option (a) 40.5 grams.

Key Concepts

Chemical Reactions

Chemical reactions are processes that involve the transformation of substances into different substances through the breaking and forming of chemical bonds.
In our exercise, a simple yet fundamental chemical reaction occurs between barium hydroxide (\(\mathrm{Ba(OH)_2}\)) and carbon dioxide (\(\mathrm{CO_2}\)).
This reaction produces barium carbonate (\(\mathrm{BaCO_3}\)) and water (\(\mathrm{H_2O}\)).
The chemical equation can be expressed as:

• \(\mathrm{Ba(OH)_2 + CO_2 \rightarrow BaCO_3 + H_2O}\)

This equation signifies a one-to-one mole relationship between barium hydroxide and carbon dioxide.
The same holds for the barium carbonate produced.
Understanding this stoichiometric relationship is crucial.
It means that for every mole of \(\mathrm{Ba(OH)_2}\) reacted, one mole of \(\mathrm{BaCO_3}\) is formed when carbon dioxide is in excess.
These stoichiometric relations help you predict the amounts of products formed from given reactants, which is the essence of stoichiometry in chemical reactions.

Molar Mass Calculations

Molar mass is a critical concept in understanding how to convert between moles and grams, a fundamental part of stoichiometry.
Molar mass is the mass of one mole of a given substance, often expressed in grams per mole (g/mol).
For barium carbonate (\(\mathrm{BaCO_3}\)), the molar mass can be calculated by adding the atomic masses of its constituent elements.

• Barium (Ba) has an atomic mass of 137 g/mol.
• Carbon (C) has an atomic mass of 12 g/mol.
• Oxygen (O) has an atomic mass of 16 g/mol, and there are three oxygen atoms.

Thus, the molar mass of \(\mathrm{BaCO_3}\) is calculated as:\[137 + 12 + (3 \times 16) = 197 \text{ g/mol}\]Understanding how to calculate molar mass is crucial as it allows you to convert between the mass of a compound and the amount in moles, enabling you to solve stoichiometry problems accurately.
With the molar mass, you can easily determine how much a certain mole of a compound weighs in grams, aiding in both theoretical and practical chemical calculations.

Mass of Compounds

Once you understand chemical reactions and how to calculate molar mass, determining the mass of compounds becomes straightforward.
In our problem, we need to calculate the mass of \(\mathrm{BaCO_3}\) formed from a given amount of \(\mathrm{Ba(OH)_2}\).
This involves using the moles of a substance and multiplying it by its molar mass.
Given that 0.205 moles of \(\mathrm{Ba(OH)_2}\) react to form 0.205 moles of \(\mathrm{BaCO_3}\), the mass of \(\mathrm{BaCO_3}\) is calculated as:\[\text{Mass} = \text{Moles} \times \text{Molar Mass}\]\[\text{Mass} = 0.205 \text{ moles} \times 197 \text{ g/mol} = 40.385 \text{ grams}\]This method illustrates how stoichiometry and molar mass calculations come together to find the mass of a compound from a given chemical reaction.
Remember, always cross-check your calculations and ensure that you multiply the correct moles with the molar mass to get an accurate result.
Such calculations bolster your understanding of chemical processes and prepare you for more complex problems.