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Chapter 18: Problem 70

A sample of iron ore weighing \(0.2792 \mathrm{~g}\) was dissolved in an excess of a dilute acid solution. All the iron was first converted to Fe(II) ions. The solution then required \(23.30 \mathrm{~mL}\) of \(0.0194 \mathrm{M} \mathrm{KMnO}_{4}\) for oxidation to Fe(III) ions. Calculate the percent by mass of iron in the ore.

Short Answer

Expert verified
The mass percent of iron in the ore is 45.17%.

Step by step solution

01

Calculate the amount of moles for KMnO4

Thevolume of the KMnO4 solution used is 23.30 mL or 0.0233 L. Using the molarity, which is moles per liter, the moles of KMnO4 can be found using the formula \[moles = molarity \times volume\]. Substituting the values gives \[moles_{KMnO4} = 0.0194 M \times 0.0233 L = 0.0004522 mol\]
02

Find the equivalent moles of Fe

In the balanced chemical equation: \[5Fe^{2+} + MnO4- + 8H+ → 5Fe^{3+} + Mn^{2+} + 4H2O\] Each molecule of KMnO4 (containing MnO4-) reacts with 5 molecules of Fe. Hence, \[moles_{Fe} = 5 \times moles_{KMnO4} = 5 \times 0.0004522 mol = 0.002261 mol\]
03

Determine the mass of iron

The molar mass of iron (Fe) is 55.845 g/mol. This means that one mole of Fe weighs 55.845g. So the mass of Fe in the sample is \[mass_{Fe} = moles \times molar~mass = 0.002261 mol \times 55.845 g/mol = 0.1262 g\]
04

Calculate the percent by mass of iron in the ore

The percent by mass of iron in the ore is computed as follows: \[ \frac{mass_{Fe}}{mass_{sample}} \times 100\% = \frac{0.1262 g}{0.2792 g} \times 100\% = 45.17\%\]

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Most popular questions from this chapter

A constant electric current flows for \(3.75 \mathrm{~h}\) through two electrolytic cells connected in series. One contains a solution of \(\mathrm{AgNO}_{3}\) and the second a solution of \(\mathrm{CuCl}_{2}\). During this time \(2.00 \mathrm{~g}\) of silver are deposited in the first cell. (a) How many grams of copper are deposited in the second cell? (b) What is the current flowing, in amperes?

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