The percentage of \(\mathrm{Fe}(\mathrm{III})\) present in iron ore \(\mathrm{Fe}_{0.93} \mathrm{O}_{1.00}\) is \((\mathrm{Fe}=56)\) (a) 94 (b) 6 (c) \(21.5\) (d) 15

Short Answer

Expert verified
Answer: 6%

Step by step solution

01

Calculate moles of Fe in the compound

Convert grams of Fe in the given sample to moles using the molar mass of Fe (56 g/mol) \(1.0\, g\, Fe * \frac{1}{56}\, \frac{mol}{g} = 0.0179\, mol\, Fe\)
02

Determine the oxidation state of Fe in the compound

Since the compound is Fe0.93O1.00, the oxidation state of Fe is given by: \(x_{Fe} * 0.93 + (-2) * 1 = 0\)
03

Calculate the oxidation state of Fe

Solve for \(x_{Fe}\) in the equation above: \(x_{Fe} = \frac{2}{0.93} = 2.15\)
04

Calculate moles of Fe(III) present in the compound

Assuming the sample only contains Fe(II) and Fe(III), the fraction of Fe(III) in the compound should be: \(Fraction\, of\, Fe(III) = 1 - \frac{2}{2.15} = 0.070\) Now, multiply the fraction of Fe(III) by the number of moles of Fe: \(0.070 * 0.0179\, mol\, Fe = 0.00125\, mol\, Fe(III)\)
05

Calculate the percentage of Fe(III) in the compound

Finally, divide moles of Fe(III) by the total moles of Fe and multiply by 100%: \(\frac{0.00125\, mol\, Fe(III)}{0.0179\, mol\, Fe} * 100\% \approx 6\%\) The percentage of Fe(III) present in the iron ore Fe0.93O1.00 is approximately 6% (Answer: b).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A mixture of \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2}\) is caused to react in a closed container to form \(\mathrm{NH}_{3} .\) The reaction ceases before either reactant has been totally consumed. At this stage, \(2.0 \mathrm{moles}\) each of \(\mathrm{N}_{2}, \mathrm{H}_{2}\) and \(\mathrm{NH}_{3}\) are present. The moles of \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2}\) present originally were, respectively, (a) 4 and 4 moles (b) 3 and 5 moles (c) 3 and 4 moles (d) 4 and 5 moles

Number of gas molecules present in \(1 \mathrm{ml}\) of gas at \(0^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\) is called Loschmidt number. Its value is about (a) \(2.7 \times 10^{19}\) (b) \(6 \times 10^{23}\) (c) \(2.7 \times 10^{22}\) (d) \(1.3 \times 10^{23}\)

A quantity of \(2.76 \mathrm{~g}\) of silver carbonate on being strongly heated yields a residue weighing \((\mathrm{Ag}=108)\) (a) \(2.16 \mathrm{~g}\) (b) \(2.48 \mathrm{~g}\) (c) \(2.32 \mathrm{~g}\) (d) \(2.64 \mathrm{~g}\)

A \(1.50 \mathrm{~g}\) sample of potassium bicarbonate having \(80 \%\) purity is strongly heated. Assuming the impurity to be thermally stable, the loss in weight of the sample, on heating, is (a) \(3.72 \mathrm{~g}\) (b) \(0.72 \mathrm{~g}\) (c) \(0.372 \mathrm{~g}\) (d) \(0.186 \mathrm{~g}\)

An amount of \(1.0 \times 10^{-3}\) moles of \(\mathrm{Ag}^{+}\) and \(1.0 \times 10^{-3}\) moles of \(\mathrm{CrO}_{4}^{2-}\) reacts together to form solid \(\mathrm{Ag}_{2} \mathrm{CrO}_{4} \cdot\) What is the amount of \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\) formed? \((\mathrm{Ag}=108, \mathrm{Cr}=52)\) (a) \(0.332 \mathrm{~g}\) (b) \(0.166 \mathrm{~g}\) (c) \(332 \mathrm{~g}\) (d) \(166 \mathrm{~g}\)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free