Question

\(\mathrm{SO}_{2}\) gas is slowly passed through an aqueous suspension containing \(12 \mathrm{~g}\) \(\mathrm{CaSO}_{3}\) till the milkiness just disappears. What amount of \(\mathrm{SO}_{2}\) would be required? (a) \(6.4\) mole (b) \(0.3\) mole (c) \(0.1 \mathrm{~mole}\) (d) \(0.2\) mole

Short answer

Answer: (c) \(0.1\, \mathrm{mol}\)

Step-by-step solution

Find the balanced chemical equation for the reaction

The reaction between \(\mathrm{CaSO}_{3}\) and \(\mathrm{SO}_{2}\) is as follows: $$\mathrm{CaSO}_{3}(s) + \mathrm{SO}_{2}(g) \rightarrow \mathrm{CaSO}_{4}(s)$$

Calculate the moles of \(\mathrm{CaSO}_{3}\)

First, find the molar mass of \(\mathrm{CaSO}_{3}\): $$\mathrm{CaSO}_{3} = \mathrm{Ca} + \mathrm{S} + (3\times \mathrm{O}) = 40.08+32.07+(3\times 16.00)= 136.22\,\mathrm{g\, mol}^{-1}$$ Given \(12\,\mathrm{g}\) of \(\mathrm{CaSO}_{3}\), calculate the moles of \(\mathrm{CaSO}_{3}\): $$\text{moles of }\mathrm{CaSO}_{3} = \frac{\text{mass}}{\text{molar mass}} = \frac{12\, \mathrm{g}}{136.22\, \mathrm{g\, mol}^{-1}} = 0.0881\, \mathrm{mol}$$

Determine the amount of \(\mathrm{SO}_{2}\) required using stoichiometry

From the balanced chemical equation, the stoichiometric ratio between \(\mathrm{CaSO}_{3}\) and \(\mathrm{SO}_{2}\) is \(1:1\) . Therefore, the amount of \(\mathrm{SO}_{2}\) required is equal to the amount of \(\mathrm{CaSO}_{3}\) present. $$\text{moles of }\mathrm{SO}_{2} \text{ required} = 0.0881\, \mathrm{mol}$$

Find the closest answer and conclude

The calculated moles of \(\mathrm{SO}_{2}\) required are \(0.0881\, \mathrm{mol}\). Looking at the given options, the closest answer is: (c) \(0.1\, \mathrm{mol}\) Therefore, the correct answer is option (c) \(0.1\,\mathrm{mol}\) of \(\mathrm{SO}_{2}\) would be required to react completely with the given amount of \(\mathrm{CaSO}_{3}\).

Key concepts

Chemical Reactions

In chemistry, a chemical reaction is a process where substances change into new and different substances. This transformation involves breaking chemical bonds in the reactants and forming new bonds in the products. In the given exercise, the reaction occurs between calcium sulfite (\(\text{CaSO}_3\)) and sulfur dioxide (\(\text{SO}_2\)) gas. When these two react, they form calcium sulfate (\(\text{CaSO}_4\)). The equation for this reaction is:

• \(\text{CaSO}_3(s) + \text{SO}_2(g) \rightarrow \text{CaSO}_4(s)\)

This reaction is a typical example of a gas reacting with a solid, resulting in a solid product. Understanding how reactants interact and form products is central to studying chemical reactions. New substances are not just mixed but chemically bonded differently.

Moles Calculation

Calculating moles is a fundamental part of solving chemical problems. It involves using the molar mass of a substance to convert between its mass and the amount in moles. The molar mass is the weight of one mole of a substance, expressed in grams per mole (g/mol).
In the exercise, we start by determining the molar mass of \(\text{CaSO}_3\):

• Calcium (\(\text{Ca}\)) = 40.08 g/mol
• Sulfur (\(\text{S}\)) = 32.07 g/mol
• Oxygen (\(\text{O}\))= 16.00 g/mol, with 3 atoms contributing (3 x 16.00)

Adding these gives a total of 136.22 g/mol for \(\text{CaSO}_3\).
Given 12 g of \(\text{CaSO}_3\), the number of moles is calculated as:

• Moles = \(\frac{12 \, \text{g}}{136.22 \, \text{g/mol}} = 0.0881 \, \text{mol}\)

This calculation is crucial as it sets the stage for determining the amount of reagent needed in the reaction.

Balancing Chemical Equations

Balancing chemical equations is an essential skill for anyone studying chemistry. It ensures that the same number of each type of atom appears on both sides of the equation, reflecting the conservation of mass principle. In this exercise, our equation:

• \(\text{CaSO}_3(s) + \text{SO}_2(g) \rightarrow \text{CaSO}_4(s)\)

We see that it's already balanced, meaning one mole of \(\text{CaSO}_3\) reacts with one mole of \(\text{SO}_2\) to produce one mole of \(\text{CaSO}_4\). The 1:1:1 ratio shows that the number of moles of \(\text{SO}_2\) needed is equal to the moles of \(\text{CaSO}_3\) available. Thus, balancing equations is not just an abstract exercise; it's integral to predicting the amounts of substances consumed and produced in a reaction.