Question

The decomposition of a certain mass of \(\mathrm{CaCO}_{3}\) gave \(11.2 \mathrm{dm}^{3}\) of \(\mathrm{CO}_{2}\) gas at STP. The mass of KOH required to completely neutralize the gas is a. \(56 \mathrm{~g}\) b. \(28 \mathrm{~g}\) c. \(42 \mathrm{~g}\) d. \(20 \mathrm{~g}\)

Short answer

The mass of KOH required is 56 g, option a.

Step-by-step solution

Understand the Reaction

First, we determine the reaction occurring here. The decomposition of calcium carbonate (\(\mathrm{CaCO}_3\)) produces carbon dioxide (\(\mathrm{CO}_2\)) and calcium oxide (\(\mathrm{CaO}\)). Additionally, \(\mathrm{CO}_2\) will react with potassium hydroxide (\(\mathrm{KOH}\)) in a neutralization reaction.The chemical equation for the decomposition is:\[ \mathrm{CaCO}_3 \rightarrow \mathrm{CaO} + \mathrm{CO}_2 \]And the neutralization reaction with KOH is:\[ \mathrm{CO}_2 + 2\mathrm{KOH} \rightarrow \mathrm{K}_2\mathrm{CO}_3} + \mathrm{H}_2\mathrm{O} \]

Calculate Moles of CO2

At standard temperature and pressure (STP), one mole of any ideal gas occupies \(22.4\mathrm{dm}^3\). Given that \(11.2\mathrm{dm}^3\) of \(\mathrm{CO}_2\) is collected, we calculate the moles of \(\mathrm{CO}_2\) using the formula:\[ \text{Moles of } \mathrm{CO}_2 = \frac{11.2\mathrm{dm}^3}{22.4\mathrm{dm}^3\text{/mol}} = 0.5\text{ moles} \]

Determine Moles of KOH Required

From the balanced neutralization equation \(\mathrm{CO}_2 + 2\mathrm{KOH} \rightarrow \mathrm{K}_2\mathrm{CO}_3 + \mathrm{H}_2\mathrm{O}\), 1 mole of \(\mathrm{CO}_2\) reacts with 2 moles of \(\mathrm{KOH}\). Thus, for 0.5 moles of \(\mathrm{CO}_2\), we require:\[ 0.5 \text{ moles of } \mathrm{CO}_2 \times 2 = 1 \text{ mole of } \mathrm{KOH} \]

Calculate Mass of KOH

The molar mass of \(\mathrm{KOH}\) is calculated as follows:\[ \mathrm{K} (39.1) + \mathrm{O} (16.0) + \mathrm{H} (1.0) = 56.1 \text{ g/mol} \]Thus, the mass of 1 mole of \(\mathrm{KOH}\) is approximately \(56 \text{ g}\). Therefore, the mass required is 56 g.

Key concepts

Decomposition Reaction

A decomposition reaction is a type of chemical reaction where a single compound breaks down into two or more simpler substances. It's like taking apart a puzzle to reveal the individual pieces. In this exercise, the compound calcium carbonate (\(\mathrm{CaCO}_3\)) undergoes decomposition.
This specific reaction can be represented with the equation:- \(\mathrm{CaCO}_3 \rightarrow \mathrm{CaO} + \mathrm{CO}_2\)

• Here, one molecule of calcium carbonate breaks apart to form one molecule of calcium oxide (\(\mathrm{CaO}\)) and one molecule of carbon dioxide (\(\mathrm{CO}_2\)).
• This type of reaction often requires an input of energy, such as heat, to proceed.

Decomposition reactions are essential in natural processes and industrial applications. Understanding them helps us learn how substances interact in different environments.

Neutralization Reaction

A neutralization reaction occurs when an acid and a base react to form water and a salt. It is a balancing act where the properties of the acid and base negate each other—a bit like bringing a seesaw to equilibrium. In this context, carbon dioxide (\(\mathrm{CO}_2\)), which acts as an acidic oxide, reacts with potassium hydroxide (\(\mathrm{KOH}\)), a base.
The balanced equation for the reaction is:- \(\mathrm{CO}_2 + 2\mathrm{KOH} \rightarrow \mathrm{K}_2\mathrm{CO}_3 + \mathrm{H}_2\mathrm{O}\)

• This means that each molecule of \(\mathrm{CO}_2\) reacts with two molecules of \(\mathrm{KOH}\) to produce one molecule of potassium carbonate (\(\mathrm{K}_2\mathrm{CO}_3\)) and one molecule of water (\(\mathrm{H}_2\mathrm{O}\)).
• Neutralization reactions are commonly involved in processes such as acid-base titrations.

Understanding neutralization reactions is crucial in fields like chemistry and environmental science, as they play a key role in maintaining chemical balance.

Molar Mass Calculation

Calculating molar mass is a fundamental skill in chemistry, allowing us to convert between mass and moles—a necessary step in many chemical calculations. The molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). To calculate it, you simply add the atomic masses of the elements in a compound, based on their abundance within one molecule of the compound.
For potassium hydroxide (\(\mathrm{KOH}\)), we calculate as follows:

• Potassium (\(\mathrm{K}\)) has an atomic mass of 39.1 g/mol.
• Oxygen (\(\mathrm{O}\)) is 16.0 g/mol.
• Hydrogen (\(\mathrm{H}\)) is 1.0 g/mol.

Adding these gives the molar mass of \(\mathrm{KOH}\) hovering around 56.1 g/mol.
This calculation aids in determining the amount of substance needed or produced in a chemical reaction. For instance, knowing the amount of \(\mathrm{KOH}\) required to neutralize \(\mathrm{CO}_2\) ensures that reactions are complete and efficient.