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Chapter 4: Problem 130

A chemical engineer determines the mass percent of iron in an ore sample by converting the Fe to \(\mathrm{Fe}^{2+}\) in acid and then titrating the \(\mathrm{Fe}^{2+}\) with \(\mathrm{MnO}_{4}^{-}\). A \(1.1081-\mathrm{g}\) ore sample was dissolved in acid and then titrated with \(39.32 \mathrm{~mL}\) of \(0.03190 \mathrm{M} \mathrm{KMnO}_{4}\). The balanced equation is \(8 \mathrm{H}^{+}(a q)+5 \mathrm{Fe}^{2+}(a q)+\mathrm{MnO}_{4}^{-}(a q) \longrightarrow\) $$ 5 \mathrm{Fe}^{3+}(a q)+\mathrm{Mn}^{2+}(a q)+4 \mathrm{H}_{2} \mathrm{O}(l) $$ Calculate the mass percent of iron in the ore.

Short Answer

Expert verified
The mass percent of iron in the ore is approximately 31.61%.

Step by step solution

01

Determine the moles of \(\text{KMnO}_{4}\)

Use the molarity and volume of \(0.03190 \text{M} \text{KMnO}_{4}\) to find the moles. Calculate it with \(n = M \cdot V\), where \(M \) is molarity and \(V \) is volume in liters. \(V = 39.32\text{ mL} = 0.03932\text{ L}\). So, the moles of \(\text{KMnO}_{4}\) is \(0.03190 \text{ M} \times 0.03932 \text{ L} = 1.254 \times 10^{-3}\text{ moles}.\)
02

Determine the moles of \( \text{Fe}^{2+} \)

Using the stoichiometry of the balanced equation, we can determine that 1 mole of \( \text{MnO}_{4}^{-}\) reacts with 5 moles of \( \text{Fe}^{2+} \). Thus, the moles of \( \text{Fe}^{2+} \) is \( 5 \times 1.254 \times 10^{-3} \text{ moles} = 6.27 \times 10^{-3} \text{ moles}. \)
03

Calculate the mass of iron in the ore sample

Use the moles of \( \text{Fe}^{2+} \) to find the mass of iron (Fe). The molar mass of Fe is 55.845 g/mol. Thus, the mass of iron is \(6.27 \times 10^{-3} \text{ moles} \times 55.845 \text{ g/mol} = 0.3503 \text{ grams}.\)
04

Calculate the mass percent of iron

Divide the mass of iron obtained by the initial mass of the ore sample and multiply by 100 to get the mass percent. Thus, the mass percent is \( \frac{0.3503 \text{ g}}{1.1081 \text{ g}} \times 100 \approx 31.61 \% \).

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Key Concepts

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

titration
Titration is a technique used to determine the concentration of a solution by reacting it with a solution of known concentration. In this exercise, titration was used to find the amount of \( \text{Fe}^{2+} \) in an iron ore sample. By adding a titrant (a reagent of known concentration), the point at which the reaction is complete is determined.
The volume of titrant used gives information about the amount of the substance being titrated. Here, \( \text{KMnO}_{4} \) of known molarity was used to titrate \( \text{Fe}^{2+} \) ions in the sample.
stoichiometry
Stoichiometry involves the calculation of reactants and products in chemical reactions. It's like the recipe of a chemical reaction, telling you how much of each ingredient you'll need and how much of each product you'll get.
In this case, the balanced reaction equation shows that 1 mole of \( \text{MnO}_{4}^{-} \) reacts with 5 moles of \( \text{Fe}^{2+} \). By using this ratio, we can convert moles of one substance to moles of another.
Knowing the moles of \( \text{KMnO}_{4} \) used, we figured out the moles of \( \text{Fe}^{2+} \) present in the sample by multiplying by the stoichiometric factor of 5.
molarity
Molarity (M) is a common measure of concentration, defined as moles of solute per liter of solution. In this exercise, the molarity of \( \text{KMnO}_{4} \) solution was given as 0.03190 M.
To find the number of moles of \( \text{KMnO}_{4} \) used in the titration, molarity is multiplied by the volume in liters: \[ n = M \times V \]
By converting 39.32 mL to liters (0.03932 L) and using the formula, we calculated the moles of \( \text{KMnO}_{4} \) used in the titration.
iron ore analysis
Iron ore analysis determines the amount of iron present in an ore sample. The sample is first dissolved in acid to convert the iron to \( \text{Fe}^{2+} \), which is then titrated with \( \text{KMnO}_{4} \). Each step of the process is crucial for accurate results.
First, accurate measurement of the ore sample and acid dissolution are essential. Then, titration with a standardized \( \text{KMnO}_{4} \) solution provides the basis for calculating the iron content.
Finally, the mass percent of iron is found by dividing the mass of iron by the total mass of the sample and multiplying by 100.

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Most popular questions from this chapter

Explain why an oxidizing agent undergoes reduction.

Thyroxine \(\left(\mathrm{C}_{15} \mathrm{H}_{11} \mathrm{I}_{4} \mathrm{NO}_{4}\right)\) is a hormone synthesized by the thyroid gland and used to control many metabolic functions in the body. A physiologist determines the mass percent of thyroxine in a thyroid extract by igniting \(0.4332 \mathrm{~g}\) of extract with sodium carbonate, which converts the iodine to iodide. The iodide is dissolved in water, and bromine and hydrochloric acid are added, which convert the iodide to iodate. (a) How many moles of iodate form per mole of thyroxine? (b) Excess bromine is boiled off and more iodide is added, which reacts as shown in the following equation: $$ \mathrm{IO}_{3}^{-}(a q)+6 \mathrm{H}^{+}(a q)+5 \mathrm{I}^{-}(a q) \longrightarrow 3 \mathrm{I}_{2}(a q)+3 \mathrm{H}_{2} \mathrm{O}(l) $$ How many moles of iodine are produced per mole of thyroxine? (Hint: Be sure to balance the charges as well as the atoms.) What are the oxidizing and reducing agents in the reaction? (c) The iodine reacts completely with \(17.23 \mathrm{~mL}\) of \(0.1000 \mathrm{M}\) thiosulfate as shown in the following unbalanced equation: $$ \mathrm{I}_{2}(a q)+\mathrm{S}_{2} \mathrm{O}_{3}^{2-}(a q) \longrightarrow \mathrm{I}^{-}(a q)+\mathrm{S}_{4} \mathrm{O}_{6}^{2-}(a q) $$ What is the mass percent of thyroxine in the thyroid extract?

When each of the following pairs of aqueous solutions is mixed, does a precipitation reaction occur? If so, write balanced molecular, total ionic, and net ionic equations: (a) Potassium carbonate + barium hydroxide (b) Aluminum nitrate \(+\) sodium phosphate

What occurs on the molecular level when an ionic compound dissolves in water?

How many total moles of ions are released when each of the following dissolves in water? (a) \(0.734 \mathrm{~mol}\) of \(\mathrm{Na}_{2} \mathrm{HPO}_{4}\) (b) \(3.86 \mathrm{~g}\) of \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) (c) \(8.66 \times 10^{20}\) formula units of \(\mathrm{NiCl}_{2}\)
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