Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 4: Problem 98

A sulfide of iron, containing \(36.5 \%\) S by mass, is heated in \(\mathrm{O}_{2}(\mathrm{g}),\) and the products are sulfur dioxide and an oxide of iron containing \(27.6 \%\) O, by mass. Write a balanced chemical equation for this reaction.

Short Answer

Expert verified
The balanced chemical equation for the reaction is \(3 FeS + 4 O_{2} \rightarrow Fe_{3}O_{4} + 3 SO_{2}\)

Step by step solution

01

Finding empirical formula of iron sulfide

The mass percentage of S in the sulfide of iron is given as \(36.5\% \). This means that the mass of S in \(100 g\) of compound is \(36.5 g\). Since the compound only contains iron and sulfur, the remaining mass must be attributed to iron: \(100 g - 36.5 g = 63.5 g\). Then, convert these masses to moles: use \(55.845 g/mol\) for iron (Fe) and \(32.06 g/mol\) for sulfur (S). \( \frac{63.5 g}{55.845 g/mol} = 1.14 mol\) (for Fe), \( \frac{36.5 g}{32.06 g/mol} = 1.14 mol\) (for S). The mole ratio of Fe to S is 1:1, thus, the empirical formula of the iron sulfide is \(FeS\).
02

Finding empirical formula of iron oxide

The mass percentage of O in the oxide of iron is given as \(27.6\% \). This means that the mass of O in \(100 g\) of compound is \(27.6 g\). The remaining mass is iron: \(100 g - 27.6 g = 72.4 g\). Convert these masses to moles: \( \frac{72.4 g}{55.845 g/mol} = 1.30 mol\) (for Fe), \( \frac{27.6 g}{16.00 g/mol} = 1.73 mol\) (for O). The mole ratio of Fe to O can be approximated as 3:4, therefore, the empirical formula of the iron oxide is \(Fe_{3}O_{4}\).
03

Balancing the chemical equation

Given the empirical formulas of the compounds and knowing that sulfur dioxide (\(SO_{2}\)) is also a product, we can write the unbalanced equation: \( FeS + O_{2} \rightarrow Fe_{3}O_{4} + SO_{2}\) Now balance for each element. Start with iron, there are 3 Fe in \(Fe_{3}O_{4}\) and 1 Fe in \(FeS\). So, multiply \(FeS\) by 3. To balance sulfur, multiply \(SO_{2}\) by 3. Now only oxygen left. There are total 8 O in the products, while there are 2 O in \(O_{2}\). So, multiply \(O_{2}\) by 4. The balanced chemical reaction becomes: \(3 FeS + 4 O_{2} \rightarrow Fe_{3}O_{4} + 3 SO_{2}\).

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.

Start your free trial

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

Under appropriate conditions, copper sulfate, potassium chromate, and water react to form a product containing \(\mathrm{Cu}^{2+},\) \(\mathrm{CrO}_{4}{^2}{^-},\) and \(\mathrm{OH}^{-}\) ions. Analysis of the compound yields \(48.7 \% \mathrm{Cu}^{2+}, 35.6 \% \mathrm{CrO}_{4}{^2}{-},\) and \(15.7 \% \mathrm{OH}^{-}\). (a) Determine the empirical formula of the compound. (b) Write a plausible equation for the reaction.

The incomplete combustion of gasoline produces \(\mathrm{CO}(\mathrm{g})\) as well as \(\mathrm{CO}_{2}(\mathrm{g}) .\) Write an equation for \((\mathrm{a})\) the complete combustion of the gasoline component octane, \(\mathrm{C}_{8} \mathrm{H}_{18}(\mathrm{l}),\) and \((\mathrm{b})\) incomplete combustion of octane with \(25 \%\) of the carbon appearing as \(\mathrm{CO}(\mathrm{g})\)

It is often difficult to determine the concentration of a species in solution, particularly if it is a biological species that takes part in complex reaction pathways. One way to do this is through a dilution experiment with labeled molecules. Instead of molecules, however, we will use fish. An angler wants to know the number of fish in a particular pond, and so puts an indelible mark on 100 fish and adds them to the pond's existing population. After waiting for the fish to spread throughout the pond, the angler starts fishing, eventually catching 18 fish. Of these, five are marked. What is the total number of fish in the pond?

A 99.8 mL sample of a solution that is \(120 \%\) KI by mass \((d=1.093 \mathrm{g} / \mathrm{mL})\) is added to \(96.7 \mathrm{mL}\) of another solution that is \(14.0 \% \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) by mass \((d=1.134 \mathrm{g} / \mathrm{mL})\) How many grams of \(\mathrm{PbI}_{2}\) should form? \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})+2 \mathrm{KI}(\mathrm{aq}) \longrightarrow \mathrm{PbI}_{2}(\mathrm{s})+2 \mathrm{KNO}_{3}(\mathrm{aq})\)

Lead nitrate and potassium iodide react in aqueous solution to form a yellow precipitate of lead iodide. In one series of experiments, the masses of the two reactants were varied, but the total mass of the two was held constant at \(5.000 \mathrm{g}\). The lead iodide formed was filtered from solution, washed, dried, and weighed. The table gives data for a series of reactions. $$\begin{array}{lll} \hline & \text { Mass of Lead } & \text { Mass of Lead } \\ \text { Experiment } & \text { Nitrate, } g & \text { lodide, } g \\ \hline 1 & 0.500 & 0.692 \\ 2 & 1.000 & 1.388 \\ 3 & 1.500 & 2.093 \\ 4 & 3.000 & 2.778 \\ 5 & 4.000 & 1.391 \\ \hline \end{array}$$ (a) Plot the data in a graph of mass of lead iodide versus mass of lead nitrate, and draw the appropriate curve(s) connecting the data points. What is the maximum mass of precipitate that can be obtained? (b) Explain why the maximum mass of precipitate is obtained when the reactants are in their stoichiometric proportions. What are these stoichiometric proportions expressed as a mass ratio, and as a mole ratio? (c) Show how the stoichiometric proportions determined in part (b) are related to the balanced equation for the reaction.
See all solutions

Recommended explanations on Chemistry Textbooks

Chemistry Branches

Read Explanation

Nuclear Chemistry

Read Explanation

Kinetics

Read Explanation

Physical Chemistry

Read Explanation

Inorganic Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation
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