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Chapter 19: Problem 63

The \(\mathrm{SO}_{2}\) present in air is mainly responsible for the phenomenon of acid rain. The concentration of \(\mathrm{SO}_{2}\) can be determined by titrating against a standard permanganate solution as follows: $$ 5 \mathrm{SO}_{2}+2 \mathrm{MnO}_{4}^{-}+2 \mathrm{H}_{2} \mathrm{O} \longrightarrow{\mathrm{SSO}_{4}^{2-}}+2 \mathrm{Mn}^{2+}+4 \mathrm{H}^{+} $$ Calculate the number of grams of \(\mathrm{SO}_{2}\) in a sample of air if \(7.37 \mathrm{~mL}\) of \(0.00800 \mathrm{M} \mathrm{KMnO}_{4}\) solution are required for the titration.

Short Answer

Expert verified
The number of grams of SO₂ in the air sample is approximately 0.00944 g.

Step by step solution

01

Understand the chemical equation

Based on the balanced chemical equation, the mole ratio of \(\mathrm{SO}_{2}\) to \(\mathrm{KMnO}_{4}\) (or \(\mathrm{MnO}_4^-\)) is 5:2. This means that 5 moles of \(\mathrm{SO}_{2}\) react with 2 moles of \(\mathrm{KMnO}_{4}\).
02

Calculate moles of KMnO₄

Molarity (\(M\)) is defined by the number of moles of solute per liter of solution. So, to calculate the number of moles of \(\mathrm{KMnO}_{4}\), multiply its molarity by the volume (in liters). The volume in this case is 7.37 mL, which is equal to 0.00737 L. So, the number of moles of \(\mathrm{KMnO}_{4}\) is \(0.00800 M * 0.00737 L = 0.00005896 mol\).
03

Calculate moles of SO₂

Use the mole ratio from step 1 (5:2) to calculate the number of moles of SO₂. The calculation will be \(\frac{5}{2} * 0.00005896 mol = 0.0001474 mol\).
04

Calculate grams of SO₂

The formula weight of SO₂ = 32.07 g (for S) + 2*16.00 g (for 2 O’s) = 64.07 g/mol. Therefore, the mass of SO₂ = 0.0001474 mol * 64.07 g/mol = 0.009444 g.

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Most popular questions from this chapter

Explain the differences between a primary galvanic cell-one that is not rechargeable- and a storage cell (for example, the lead storage battery), which is rechargeable.

The hydrogen-oxygen fuel cell is described in Section 19.6 . (a) What volume of \(\mathrm{H}_{2}(g)\), stored at \(25^{\circ} \mathrm{C}\) at a pressure of 155 atm, would be needed to run an electric motor drawing a current of \(8.5 \mathrm{~A}\) for \(3.0 \mathrm{~h} ?\) (b) What volume (liters) of air at \(25^{\circ} \mathrm{C}\) and \(1.00 \mathrm{~atm}\) will have to pass into the cell per minute to run the motor? Assume that air is 20 percent \(\mathrm{O}_{2}\) by volume and that all the \(\mathrm{O}_{2}\) is consumed in the cell. The other components of air do not affect the fuel-cell reactions. Assume ideal gas behavior.

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