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Chapter 8: Problem 92

\(\mathrm{SO}_{2}\) 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

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

Step by step solution

01

Find the balanced chemical equation for the reaction

The reaction between \(\mathrm{CaSO}_{3}\) and \(\mathrm{SO}_{2}\) is as follows: $$\mathrm{CaSO}_{3}(s) + \mathrm{SO}_{2}(g) \rightarrow \mathrm{CaSO}_{4}(s)$$
02

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 \(\mathrm{CaSO}_{3}\): $$\text{moles of }\mathrm{CaSO}_{3} = \frac{\text{mass}}{\text{molar mass}} = \frac{12\, \mathrm{g}}{136.22\, \mathrm{g\, mol}^{-1}} = 0.0881\, \mathrm{mol}$$
03

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. $$\text{moles of }\mathrm{SO}_{2} \text{ required} = 0.0881\, \mathrm{mol}$$
04

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 \(\mathrm{CaSO}_{3}\).

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Most popular questions from this chapter

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