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Chapter 23: Problem 64

A 0.589 g sample of pyrolusite ore (impure \(\mathrm{MnO}_{2}\) ) is treated with \(1.651 \mathrm{g}\) of oxalic acid \(\left(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4} \cdot 2 \mathrm{H}_{2} \mathrm{O}\right)\) in an acidic medium (reaction 1). Following this, the excess oxalic acid is titrated with \(30.06 \mathrm{mL}\) of \(0.1000 \mathrm{M}\) \(\mathrm{KMnO}_{4}\) (reaction 2). What is the mass percent of \(\mathrm{MnO}_{2}\) in the pyrolusite? The following equations are neither complete nor balanced. (1) \(\quad \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(\mathrm{aq})+\mathrm{MnO}_{2}(\mathrm{s}) \longrightarrow \mathrm{Mn}^{2+}(\mathrm{aq})+\mathrm{CO}_{2}(\mathrm{g})\) (2) \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(\mathrm{aq})+\mathrm{MnO}_{4}^{-}(\mathrm{aq}) \longrightarrow \mathrm{Mn}^{2+}(\mathrm{aq})+\mathrm{CO}_{2}(\mathrm{g})\)

Short Answer

Expert verified
The mass percent of MnO2 in the pyrolusite is 41.22%.

Step by step solution

01

Balance the reaction equations

Reaction 1: 2H2C2O4(aq) + MnO2(s) -> Mn2+(aq) + CO2(g) + 2H2O(l) + 2e-. This reaction involves reduction. Reaction 2: 5H2C2O4(aq) + 2MnO4-(aq) -> 10CO2(g) + 8H2O(l) + 2Mn2+(aq). This reaction involves oxidation.
02

Calculate the moles of KMnO4

KMnO4 reacts with the excess oxalic acid. Given the molarity (0.1000 M), and volume (30.06 mL or 0.03006 L), the moles can be determined: mol KMnO4 = Molarity * Volume = 0.1000 M * 0.03006 L = 0.003006 mol KMnO4.
03

Determine the moles of H2C2O4 reacted with KMnO4

From the balanced equation 5H2C2O4 -> 2 KMnO4, every 5 moles of oxalic acid react with 2 moles of KMnO4. Therefore, the moles of oxalic acid that reacted with KMnO4 are: mol H2C2O4 = (mol KMnO4 * 5) / 2 = 0.003006 * 5 / 2 = 0.007515 mol H2C2O4.
04

Calculate the total moles of H2C2O4

The total moles of oxalic acid are given by the mass (in g) divided by the molar mass. Given mass is 1.651g and molar mass is 126.07 g/mol, therefore the moles = mass / molar mass = 1.651g / 126.07g/mol = 0.0131 mol H2C2O4.
05

Determine the moles of H2C2O4 reacted with MnO2

The moles of oxalic acid reacted with MnO2 are given by the difference between total moles of oxalic acid and those reacted with KMnO4: mol H2C2O4 = total mol H2C2O4 - mol reacted with KMnO4 = 0.0131 - 0.007515 = 0.005585 mol H2C2O4.
06

Calculate the moles and mass of MnO2

From the balanced equation, every 2 moles of oxalic acid react with 1 mole of MnO2. Therefore, the moles of MnO2 that reacted are given by: mol MnO2 = mol H2C2O4 / 2 = 0.005585 / 2 = 0.0027925 mol MnO2. The mass of MnO2 is given by multiply this with its molar mass: mass = mol MnO2 * Molar mass MnO2 = 0.0027925 mol * 86.94 g/mol = 0.2428 g.
07

Calculate the mass percent of MnO2

The mass percent of MnO2 is given by the mass of MnO2 reacted divided by the total mass of the sample multiplied by 100%: MnO2 mass percent = (mass MnO2/ sample mass) * 100% = (0.2428 g / 0.589 g) * 100% = 41.22%

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Most popular questions from this chapter

The only important compounds of \(\mathrm{Ag}(\mathrm{II})\) are \(\mathrm{AgF}_{2}\) and AgO. Why would you expect these two compounds to be stable, but not other silver(II) compounds such as \(\mathrm{AgCl}_{2}, \mathrm{AgBr}_{2},\) and \(\mathrm{AgS} ?\)

Arrange the following species according to the number of unpaired electrons they contain, starting with the one that has the greatest number: \(\mathrm{Fe}, \mathrm{Sc}^{3+}, \mathrm{Ti}^{2+}\) \(\mathrm{Mn}^{4+}, \mathrm{Cr}, \mathrm{Cu}^{2+}\).

All of the following elements have an ion displaying the +6 oxidation state except (a) Mo; (b) Cr; (c) Mn; (d) \(\mathrm{V} ;\) (e) S.

Why is the number of common oxidation states for the elements at the beginning and those at the end of the first transition series less than for elements in the middle of the series?

Show that under the following conditions, \(\mathrm{Ba}^{2+}(\mathrm{aq})\) can be separated from \(\mathrm{Sr}^{2+}(\mathrm{aq})\) and \(\mathrm{Ca}^{2+}(\) aq) by precipitating \(\mathrm{BaCrO}_{4}(\mathrm{s})\) with the other ions remaining in solution: $$\begin{aligned} &\left[\mathrm{Ba}^{2+}\right]=\left[\mathrm{Sr}^{2+}\right]=\left[\mathrm{Ca}^{2+}\right]=0.10 \mathrm{M}\\\ &\left[\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right]=\left[\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right]=1.0 \mathrm{M}\\\ &\left[\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\right]=0.0010 \mathrm{M}\\\ &\mathrm{K}_{\mathrm{sp}}\left(\mathrm{BaCrO}_{4}\right)=1.2 \times 10^{-10}\\\ &K_{\mathrm{sp}}\left(\mathrm{SrCrO}_{4}\right)=2.2 \times 10^{-5} \end{aligned}$$ Use data from this and previous chapters, as necessary.
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