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Chapter 8: Problem 11

Iron pyrites are converted to iron oxide by the process of

Short Answer

Expert verified
Answer: The balanced chemical equation for the conversion of iron pyrites (FeS2) to iron oxide (Fe2O3) is 2FeS2 + O2 -> Fe2O3 + 2SO2.

Step by step solution

01

Write down the given reactant and product

Write down the initial reactant, iron pyrites (FeS2), and the final product, iron oxide (Fe2O3).
02

Balance the elements other than oxygen

In the chemical equation, balance the elements other than oxygen. We notice here that there are two iron atoms on the product side and only one iron atom on the reactant side. So, we need to balance the iron atoms by writing: 2FeS2 -> Fe2O3
03

Determine the oxidation states

In order to find out how the sulfur in iron pyrites is converted to oxygen in iron oxide, we must determine the oxidation states for each element in the reactants and products. Using the rules of oxidation states, we can write: FeS2: Fe(II), S(-II) Fe2O3: Fe(III), O(-II)
04

Identify the redox process

From the oxidation states, we see that iron is oxidized from Fe(II) to Fe(III). In this reaction, sulfur is also oxidized from S(-II) to O(-II). The oxygen in the air is reduced to O(-II) in the product. Thus, the overall reaction involves iron being oxidized and oxygen being reduced.
05

Balance the redox reaction

Now, we need to balance the redox reaction by adjusting the stoichiometric coefficients of the reactants and products: 2FeS2 + O2 -> Fe2O3 + 2SO2 In this balanced redox reaction: - Iron pyrites are oxidized to iron oxide. - Oxygen in the air (O2) is reduced to form sulfur dioxide (SO2).
06

Summarize the conversion process

To summarize the process of converting iron pyrites to iron oxide: Step 1: Write down the given reactant and product. Step 2: Balance the elements other than oxygen. Step 3: Determine the oxidation states for each element. Step 4: Identify the redox process. Step 5: Balance the redox reaction. Step 6: Summarize the conversion process. The balanced chemical equation for the conversion of iron pyrites (FeS2) to iron oxide (Fe2O3) is: 2FeS2 + O2 -> Fe2O3 + 2SO2

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Most popular questions from this chapter

Hydrogen cyanide, HCN, is prepared from ammonia, air and natural gas \(\left(\mathrm{CH}_{4}\right)\) by the following process. \(2 \mathrm{NH}_{3}(\mathrm{~g})+3 \mathrm{O}_{2}(\mathrm{~g})+2 \mathrm{CH}_{4}(\mathrm{~g}) \stackrel{\mathrm{Pt}}{\longrightarrow}\) \(2 \mathrm{HCN}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) If a reaction vessel contains \(11.5 \mathrm{~g} \mathrm{NH}_{3}\), \(10.0 \mathrm{~g} \mathrm{O}_{2}\), and \(10.5 \mathrm{~g} \mathrm{CH}_{4}\), what is the maximum mass, in grams, of hydrogen cyanide that could be made, assuming the reaction goes to completion? (a) \(18.26 \mathrm{~g}\) (b) \(5.625 \mathrm{~g}\) (c) \(17.72 \mathrm{~g}\) (d) \(16.875 \mathrm{~g}\)

An aqueous solution has urea and glucose in mass ratio \(3: 1\). If the mass ratio of water and glucose in the solution is \(10: 1\), then the mole fraction of glucose in the solution is (a) \(\frac{1}{110}\) (b) \(\frac{9}{110}\) (c) \(\frac{3}{110}\) (d) \(\frac{100}{110}\)

When \(20 \mathrm{~g} \mathrm{Fe}_{2} \mathrm{O}_{3}\) is reacted with \(50 \mathrm{~g}\) of \(\mathrm{HCl}, \mathrm{FeCl}_{3}\) and \(\mathrm{H}_{2} \mathrm{O}\) are formed. The amount of unreacted \(\mathrm{HCl}\) is \((\mathrm{Fe}=56)\) (a) \(27.375 \mathrm{~g}\) (b) \(22.625 \mathrm{~g}\) (c) \(30 \mathrm{~g}\) (d) \(4.75 \mathrm{~g}\)

What mass of solid ammonium carbonate \(\mathrm{H}_{2} \mathrm{NCOONH}_{4}\), when vaporized at \(273^{\circ} \mathrm{C}\), will have a volume of \(8.961\) at \(760 \mathrm{~mm}\) of pressure. Assume that the solid completely decomposes as \(\mathrm{H}_{2} \mathrm{NCOONH}_{4}(\mathrm{~s}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g})+2 \mathrm{NH}_{3}(\mathrm{~g})\) (a) \(15.6 \mathrm{~g}\) (b) \(5.2 \mathrm{~g}\) (c) \(46.8 \mathrm{~g}\) (d) \(7.8 \mathrm{~g}\)

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
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