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 \(\mathrm{Fe}(\mathrm{II})\) 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
After performing the above calculations, this will provide the percent by mass of iron in the ore sample.

Step by step solution

01

Determine the moles of KMnO4

Conversion of volume of solution to number of moles uses the formula: \(n = MV\), where \(n\) is the number of moles, \(M\) is the molarity and \(V\) is the volume, in liters. Therefore, \(n_{KMnO4} = 0.0194 \times 23.30\times 10^{-3}\) mol.
02

Find number of the moles of Iron

Using the redox half reactions, the stoichiometry shows that 1 mole of KMnO4 reacts with 5 moles of Fe(II). Therefore, the number of moles of Fe in the sample calculated as : \(n_{Fe} = n_{KMnO4} \times 5\).
03

Calculate the weight of Iron in the sample

Multiply the moles of iron by the molar mass of iron which is \(55.845 g/mol\) to get the mass: weight of Fe = \(n_{Fe} \times 55.845 g/mol\)
04

Calculate the percentage of Iron

Divide the weight of Fe by the total mass of the sample and multiply by 100 to find the percentage by mass of iron in the sample: \[ \% Fe = \left(\frac{weight \, of \, Fe}{0.2792} \right) \times 100 \%\]

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Most popular questions from this chapter

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