/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 54 If steel wool (iron) is heated u... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 54

If steel wool (iron) is heated until it glows and is placed in a bottle containing pure oxygen, the iron reacts spectacularly to produce iron(III) oxide. $$\mathrm{Fe}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{Fe}_{2} \mathrm{O}_{3}(s)$$ If \(1.25 \mathrm{g}\) of iron is heated and placed in a bottle containing 0.0204 mol of oxygen gas, what mass of iron(III) oxide is produced?

Short Answer

Expert verified
The limiting reactant is iron, and 1.79 grams of iron(III) oxide are produced when 1.25 g of iron reacts with 0.0204 mol of oxygen gas.

Step by step solution

01

Calculate the moles of iron and oxygen

First, we need to find the moles of iron and oxygen gas. The moles of iron can be calculated by dividing the mass of iron by its molar mass: Moles of iron = Mass of iron / Molar mass of iron Molar mass of iron (Fe) = 55.85 g/mol Moles of iron = 1.25 g / 55.85 g/mol = 0.0224 mol We are given the moles of oxygen gas, which is 0.0204 mol.
02

Determine the limiting reactant

Compare the molar ratio of the reactants with the stoichiometry of the reaction to determine the limiting reactant. The balanced chemical equation is: \(4 \mathrm{Fe}(s) + 3\mathrm{O}_{2}(g) \rightarrow 2\mathrm{Fe}_{2}\mathrm{O}_{3}(s)\) Divide the moles of each reactant by their stoichiometric coefficients: Moles of iron / 4 = 0.0224 mol / 4 = 0.0056 Moles of oxygen / 3 = 0.0204 mol / 3 = 0.0068 Iron has the smaller ratio (0.0056), so iron is the limiting reactant.
03

Calculate the moles of iron(III) oxide produced

Using the stoichiometry of the reaction, we can determine the moles of iron(III) oxide produced by multiplying the moles of the limiting reactant (iron) by the stoichiometric ratio of iron(III) oxide to iron: Moles of iron(III) oxide = Moles of iron 脳 (2 mol iron(III) oxide / 4 mol iron) Moles of iron(III) oxide = 0.0224 mol 脳 (2/4) = 0.0112 mol
04

Calculate the mass of iron(III) oxide produced

Finally, we will convert the moles of iron(III) oxide to mass using the molar mass of iron(III) oxide: Mass of iron(III) oxide = Moles of iron(III) oxide 脳 Molar mass of iron(III) oxide The molar mass of iron(III) oxide is the sum of the molar masses of two iron atoms and three oxygen atoms: Molar mass of iron(III) oxide = 2 脳 55.85 g/mol + 3 脳 16 g/mol = 159.70 g/mol Mass of iron(III) oxide = 0.0112 mol 脳 159.70 g/mol = 1.79 g So, 1.79 grams of iron(III) oxide are produced.

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 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Limiting Reactant
In any chemical reaction, the limiting reactant plays a crucial role as it determines the maximum amount of product that can be formed. It is the reactant that is entirely consumed first, thereby halting the reaction. Identifying the limiting reactant is essential for accurate stoichiometry calculations.

To find the limiting reactant, compare the mole ratio of each reactant with the stoichiometric coefficients from the balanced chemical equation:
  • Divide the calculated moles of each reactant by their respective coefficients in the balanced equation.
  • The reactant that has the smallest resultant value is the limiting reactant.
In the example provided, iron is the limiting reactant because it has a smaller ratio compared to oxygen when divided by their coefficients from the equation. This means that the amount of iron limits how much iron(III) oxide can be produced.
Chemical Reactions
Chemical reactions involve the transformation of reactants into products through the breaking and forming of chemical bonds. In our example, iron and oxygen react to form iron(III) oxide, which is a classic example of a synthesis reaction.

The balanced chemical equation is vital because it shows the precise ratio in which the reactants combine. In the reaction:\[ 4 ext{Fe}(s) + 3 ext{O}_2(g) ightarrow 2 ext{Fe}_2 ext{O}_3(s) \]
  • It indicates that 4 moles of iron react with 3 moles of oxygen to produce 2 moles of iron(III) oxide.
  • Understanding this ratio helps in calculating the theoretical yield of products in mole or mass terms, based on the amounts of limiting reactant available.
Chemical equations not only show the substances involved but also highlight the conservation of mass, important for subsequent mole and mass calculations.
Mole Calculations
Mole calculations are fundamental in stoichiometry as they link the macroscopic world to the atomic scale. Knowing how to calculate the number of moles from given masses or volumes helps predict the amount of product formed in reactions.

The mole is a bridge between the atomic world and the macro world, defining amounts in terms of Avogadro's number, which is approximately \(6.022 \times 10^{23}\) entities per mole.

Here鈥檚 how mole calculations work in our example:
  • To find the moles, divide the mass of a substance by its molar mass. For instance, the moles of iron were calculated by dividing its mass by its molar mass (55.85 g/mol).
  • Similarly, if a reactant is in gaseous form, its moles might be given directly, as with the oxygen gas.
By calculating moles, students can determine how much of a product will form under given conditions, leading to practical applications in laboratory settings and industrial processes.

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

One method for chemical analysis involves finding some reagent that will precipitate the species of interest. The mass of the precipitate is then used to determine what mass of the species of interest was present in the original sample. For example, calcium ion can be precipitated from solution by addition of sodium oxalate. The balanced equation is $$\mathrm{Ca}^{2+}(a q)+\mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(a q) \rightarrow \mathrm{CaC}_{2} \mathrm{O}_{4}(s)+2 \mathrm{Na}^{+}(a q)$$ Suppose a solution is known to contain approximately 15 g of calcium ion. Show by calculation whether the addition of a solution containing \(15 \mathrm{g}\) of sodium oxalate will precipitate all of the calcium from the sample.

The halogen elements are so reactive that the halides of many metals can be prepared by the direct combination of the elements. For example, iron(III) chloride can be prepared by the following reaction: $$2 \mathrm{Fe}(s)+3 \mathrm{Cl}_{2}(g) \rightarrow 2 \mathrm{FeCl}_{3}(s)$$ Calculate the mass of \(\mathrm{FeCl}_{3}\) that is formed if \(12.4 \mathrm{g}\) of iron reacts completely.

The compound sodium thiosulfate pentahydrate, \(\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3} \cdot 5 \mathrm{H}_{2} \mathrm{O},\) is important commercially to the photography business as "hypo," because it has the ability to dissolve unreacted silver salts from photographic film during development. Sodium thiosulfate pentahydrate can be produced by boiling elemental sulfur in an aqueous solution of sodium sulfite. $$\mathrm{S}_{8}(s)+\mathrm{Na}_{2} \mathrm{SO}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3} \cdot 5 \mathrm{H}_{2} \mathrm{O}(s)$$ (unbalanced) What is the theoretical yield of sodium thiosulfate pentahydrate when \(3.25 \mathrm{g}\) of sulfur is boiled with 13.1 g of sodium sulfite? Sodium thiosulfate pentahydrate is very soluble in water. What is the percent yield of the synthesis if a student doing this experiment is able to isolate (collect) only \(5.26 \mathrm{g}\) of the product?

Ammonium nitrate has been used as a high explosive because it is unstable and decomposes into several gaseous substances. The rapid expansion of the gaseous substances produces the explosive force. $$\mathrm{NH}_{4} \mathrm{NO}_{3}(s) \rightarrow \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)$$ Calculate the mass of each product gas if \(1.25 \mathrm{g}\) of ammonium nitrate reacts.

For each of the following unbalanced chemical equations, suppose that exactly \(5.00 \mathrm{g}\) of each reactant is taken. Determine which reactant is limiting, and calculate what mass of each product is expected (assuming that the limiting reactant is completely consumed). a. \(\mathrm{S}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{SO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)\) b. \(\operatorname{MnO}_{2}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(l) \rightarrow \mathrm{Mn}\left(\mathrm{SO}_{4}\right)_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l)\) c. \(\mathrm{H}_{2} \mathrm{S}(g)+\mathrm{O}_{2}(g) \rightarrow \mathrm{SO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)\) d. \(\mathrm{AgNO}_{3}(a q)+\mathrm{Al}(s) \rightarrow \mathrm{Ag}(s)+\mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}(a q)\)
See all solutions

Recommended explanations on Chemistry Textbooks

The Earths Atmosphere

Read Explanation

Nuclear Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Chemical Analysis

Read Explanation

Making Measurements

Read Explanation

Chemical Reactions

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.