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Chapter 4: Problem 73

Limonite, an ore of iron, is brought into solution in acidic medium and titrated with \(\mathrm{KMnO}_{4}\). The unbalanced equation for the reaction is $$ \mathrm{MnO}_{4}^{-}(a q)+\mathrm{Fe}^{2+}(a q) \longrightarrow \mathrm{Fe}^{3+}(a q)+\mathrm{Mn}^{2+}(a q) $$ It is found that a \(1.000-\mathrm{g}\) sample of the ore requires \(75.52 \mathrm{~mL}\) of \(0.0205 \mathrm{M}\) \(\mathrm{KMnO}_{4}\). What is the percent of Fe in the sample?

Short Answer

Expert verified
Answer: The percent of Fe in the limonite sample is 69.22%.

Step by step solution

01

Balance the chemical equation

To balance the chemical equation, we need 8 moles of H2O on the right side and 5 moles of H+ ions on the left side. The balanced equation becomes: $$ 5 \mathrm{H}_{2}\mathrm{O}(l)+\mathrm{MnO}_{4}^{-}(a q)+8 \mathrm{Fe}^{2+}(a q) \longrightarrow 5 \mathrm{H}^{+}(a q)+8 \mathrm{Fe}^{3+}(a q)+\mathrm{Mn}^{2+}(a q) $$
02

Calculate moles of KMnO4 used

To calculate the moles of KMnO4 used in the titration, we start with the given volume and concentration: Moles of KMnO4 = \(\text{volume (L)} \times \text{concentration (M)}\) Moles of KMnO4 = \((75.52\times10^{-3})\, \text{L} \times 0.0205\,\text{M} = 1.5486 \times10^{-3}\,\text{moles}\)
03

Calculate moles of Fe in the sample

Using the balanced equation and stoichiometry, we can find the moles of Fe in the sample: $$ \frac{1.5486\times10^{-3}\,\text{moles KMnO}_4}{1} \times \frac{8\,\text{moles Fe}^2+}{1\,\text{mole KMnO}_4} = 1.239\times10^{-2}\,\text{moles Fe}^2+ $$
04

Calculate the mass and percent of Fe in the sample

Now that we have the moles of Fe, we can convert it to mass using the molar mass of Fe: Mass of Fe = moles of Fe × molar mass of Fe Mass of Fe = \(1.239\times10^{-2}\,\text{moles} \times 55.85\, \frac{\text{g}}{\text{mole}} = 0.6922\,\text{g}\) Now, we can find the percent of Fe in the sample: Percent of Fe = \(\frac{\text{Mass of Fe}}{\text{Mass of limonite sample}}\times100\) Percent of Fe = \(\frac{0.6922\,\text{g}}{1.000\,\text{g}}\times100 = 69.22\%\) Therefore, the percent of Fe in the limonite sample is 69.22%.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Stoichiometry in Chemistry
Stoichiometry is a section of chemistry that involves using relationships from the balanced chemical equations to calculate the quantities of reactants and products. It's like a recipe for chemistry where you need to know the precise amounts of ingredients you need and how much you'll end up with.

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Most popular questions from this chapter

A solution of potassium permanganate reacts with oxalic acid, \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) to form carbon dioxide and solid manganese(IV) oxide \(\left(\mathrm{MnO}_{2}\right)\). (a) Write a balanced net ionic equation for the reaction. (b) If \(20.0 \mathrm{~mL}\) of \(0.300 M\) potassium permanganate is required to react with \(13.7 \mathrm{~mL}\) of oxalic acid, what is the molarity of the oxalic acid? (c) What is the mass of manganese(IV) oxide formed?

Three students titrate different samples of the same solution of \(\mathrm{HCl}\) to obtain its molarity. Below are their data. Student \(\mathrm{A}: \quad 20.00 \mathrm{~mL} \mathrm{HCl}+20.00 \mathrm{~mL} \mathrm{H}_{2} \mathrm{O}\) \(0.100 \mathrm{M} \mathrm{NaOH}\) used to titrate to the equivalence point Student B: \(\quad 20.00 \mathrm{~mL} \mathrm{HCl}+40.00 \mathrm{~mL} \mathrm{H}_{2} \mathrm{O}\) \(0.100 \mathrm{M} \mathrm{NaOH}\) used to titrate to the equivalence point Student C: \(\quad 20.00 \mathrm{~mL} \mathrm{HCl}+20.00 \mathrm{~mL} \mathrm{H}_{2} \mathrm{O}\) \(0.100 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) used to titrate to the equivalence point. All the students calculated the molarities correctly. Which (if any) of the following statements are true? (a) The molarity calculated by \(A\) is half that calculated by \(B\). (b) The molarity calculated by \(\mathrm{A}\) is equal to that calculated by \(\mathrm{C}\). (c) The molarity calculated by B is twice that calculated by C. (d) The molarity calculated by \(\mathrm{A}\) is twice that calculated by \(\mathrm{B}\). (e) The molarity calculated by \(\mathrm{A}\) is equal to that calculated by \(\mathrm{B}\). Challenge Problems

Laundry bleach is a solution of sodium hypochlorite (NaClO). To determine the hypochlorite (ClO \(^{-}\) ) content of bleach (which is responsible for its bleaching action), sulfide ion is added in basic solution. The balanced equation for the reaction is $$ \mathrm{ClO}^{-}(a q)+\mathrm{S}^{2-}(a q)+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{Cl}^{-}(a q)+\mathrm{S}(s)+2 \mathrm{OH}^{-}(a q) $$ The chloride ion resulting from the reduction of HClO is precipitated as \(\mathrm{AgCl}\). When \(50.0 \mathrm{~mL}\) of laundry bleach \(\left(d=1.02 \mathrm{~g} / \mathrm{cm}^{3}\right)\) is treated as described above, \(4.95 \mathrm{~g}\) of \(\mathrm{AgCl}\) is obtained. What is the mass percent of \(\mathrm{NaClO}\) in the bleach?

For an acid-base reaction, what is the reacting species (the ion or molecule that appears in the chemical equation) in the following bases? (a) barium hydroxide (b) trimethylamine \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{~N}\) (c) aniline, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\) (d) sodium hydroxide

Write net ionic equations for the formation of (a) a precipitate when solutions of magnesium nitrate and potassium hydroxide are mixed. (b) two different precipitates when solutions of silver(I) sulfate and barium chloride are mixed.
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