/*! 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 113 An iron bar weighed \(664 \mathr... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 113

An iron bar weighed \(664 \mathrm{~g}\). After the bar had been standing in moist air for a month, exactly one-eighth of the iron turned to rust \(\left(\mathrm{Fe}_{2} \mathrm{O}_{3}\right)\). Calculate the final mass of the iron bar and rust.

Short Answer

Expert verified
The final mass is 699.57 g.

Step by step solution

01

Understand the Problem

We have an iron bar weighing 664 g, and one-eighth of it has turned into rust, which is iron oxide, noted as \(\mathrm{Fe}_2\mathrm{O}_3\). Our task is to find the final mass of the iron bar including the rust.
02

Calculate the Mass of Iron that Turned to Rust

First, determine the mass of the iron that turned to rust by dividing the original mass of the iron bar by 8. \[ \text{Mass of iron turned to rust} = \frac{664}{8} = 83 \text{ g} \]
03

Determine the Mass of Rust Formed

Realize that rust forms from iron and oxygen. The molar mass of iron (Fe) is 56 g/mol and \(\mathrm{Fe}_2\mathrm{O}_3\) is (56*2 + 16*3) = 160 g/mol. Use stoichiometry to find the mass of rust from 83 g of iron. If 1 mole of iron (56 g) converts to part of a mole of rust, 2 moles of iron (112 g) will yield 160 g of \(\mathrm{Fe}_2\mathrm{O}_3\). The proportion is: \[ \frac{160 ext{ g rust}}{112 ext{ g iron}} = \frac{x ext{ g rust}}{83 ext{ g iron}} \] Solve for \(x\): \[ x = \frac{160}{112} \times 83 = 118.57 \text{ g of rust} \]
04

Calculate the Final Mass

Add the mass of the remaining iron and the rust. The remaining iron is 664 g - 83 g = 581 g. Thus the total mass after rusting is: \[ \text{Total mass} = 581 ext{ g (remaining iron)} + 118.57 ext{ g (rust)} = 699.57 ext{ g} \]
05

Conclusion

The final mass of the iron bar, including the rust, is 699.57 g. This accounts for the original iron that did not rust and the additional mass from the oxygen that combined with some of the iron to form rust.

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.

Understanding Chemical Reactions
Chemical reactions are processes in which reactants transform into products. Here, the chemical reaction in focus is the rusting of iron. Rusting occurs when iron reacts with oxygen in the presence of moisture to form iron oxide (\(\mathrm{Fe}_2\mathrm{O}_3\)). This process involves the combination of iron and oxygen atoms to form a new compound.
Chemical reactions are often represented by balanced chemical equations, which describe the reactants and products involved. For the rusting of iron, the simplified equation is:\[4\mathrm{Fe} + 3\mathrm{O}_2 \rightarrow 2\mathrm{Fe}_2\mathrm{O}_3\]Meaning, four iron atoms react with three oxygen molecules to yield two units of iron oxide. This helps us understand how much reactant is needed and what is produced.
Understanding Molar Mass
Molar mass is the mass of one mole of a substance, usually expressed in grams per mole. This concept is essential in stoichiometry to test relationships in a chemical reaction.
For example:
  • The molar mass of iron (Fe) is 56 g/mol.
  • The molar mass of diatomic oxygen (\(\mathrm{O}_2\)) is 32 g/mol (since 16 g/mol per oxygen atom).
  • The molar mass of iron oxide (\(\mathrm{Fe}_2\mathrm{O}_3\)) is 160 g/mol, calculated as (56 g/mol \(\times 2\) for iron) plus (16 g/mol \(\times 3\) for oxygen).
Molar mass allows us to convert between grams and moles, facilitating the determination of how much product is formed from given quantities of reactants.
What is Iron Oxide?
Iron oxide, specifically \(\mathrm{Fe}_2\mathrm{O}_3\), is a reddish-brown compound formed from iron and oxygen. It's commonly known as rust and typically forms when iron is exposed to air and moisture over time.
Key Characteristics:
  • Composed of two iron atoms and three oxygen atoms.
  • Inefficient as a protective layer for iron since it can flake off, exposing new layers to corrosion.
  • Used industrially as a pigment and in various applications such as in ceramics and metallurgy.
Understanding iron oxide is crucial in addressing how metals like iron degrade and are preserved over time.
Performing Mass Calculations
Mass calculations are an essential part of working through chemical reactions, determining how much of each substance is involved. In our exercise, we are concerned with the initial, rusted, and final mass of the iron.
Steps in Mass Calculation:
  • Start by determining what portion of the iron bar rusted, here one-eighth turning to rust equals 83 g.
  • Using stoichiometry and the known molar masses, calculate the mass of iron oxide formed from the rusted portion, here found to be 118.57 g.
  • Add the remaining iron to this rust mass for the total mass, resulting in 699.57 g as the final mass.
These steps illustrate the concept of conservation of mass, where mass is neither created nor destroyed but redistributed in chemical reactions.

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

When heated, lithium reacts with nitrogen to form lithium nitride: $$ 6 \mathrm{Li}(s)+\mathrm{N}_{2}(g) \stackrel{\Delta}{\longrightarrow} 2 \mathrm{Li}_{3} \mathrm{~N}(s) $$ What is the theoretical yield of \(\mathrm{Li}_{3} \mathrm{~N}\) in grams when \(12.3 \mathrm{~g}\) of \(\mathrm{Li}\) is heated with \(33.6 \mathrm{~g}\) of \(\mathrm{N}_{2}\) ? If the actual yield of \(\mathrm{Li}_{2} \mathrm{~N}\) is \(5.89 \mathrm{~g}\), what is the percent vield of the reaction?

Leaded gasoline contains an additive to prevent engine "knocking." On analysis, the additive compound is found to contain carbon, hydrogen, and lead (Pb) (hence, "leaded gasoline"). When \(51.36 \mathrm{~g}\) of this compound is burned in an apparatus such as that shown in Figure \(3.5,55.90 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) and \(28.61 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\) are produced. Determine the empirical formula of the gasoline additive. Because of its detrimental effect on the environment, the original lead additive has been replaced in recent years by methyl tert-butyl ether (a compound of \(\mathrm{C}, \mathrm{H},\) and \(\mathrm{O}\) ) to enhance the performance of gasoline. (As of \(1999,\) this compound is also being phased out because of its contamination of drinking water.) When \(12.1 \mathrm{~g}\) of the compound is burned in an apparatus like the one shown in Figure \(3.5,30.2 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) and \(14.8 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\) are formed. What is the empirical formula of this compound?

Industrially, nitric acid is produced by the Ostwald process represented by the following equations: $$ \begin{aligned} 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \longrightarrow & 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \\ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{NO}_{2}(g) \\ 2 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) & \longrightarrow \mathrm{HNO}_{3}(a q)+\mathrm{HNO}_{2}(a q) \end{aligned} $$ What mass of \(\mathrm{NH}_{3}\) (in grams) must be used to produce 1.00 ton of \(\mathrm{HNO}_{3}\) by the Ostwald process, assuming an 80 percent yield in each step \((1\) ton \(=2000 \mathrm{lb} ;\) $$ 1 \mathrm{lb}=453.6 \mathrm{~g}) ? $$

Nitroglycerin \(\left(\mathrm{C}_{3} \mathrm{H}_{5} \mathrm{~N}_{3} \mathrm{O}_{9}\right)\) is a powerful explosive. Its decomposition may be represented by $$ 4 \mathrm{C}_{3} \mathrm{H}_{5} \mathrm{~N}_{3} \mathrm{O}_{9} \longrightarrow 6 \mathrm{~N}_{2}+12 \mathrm{CO}_{2}+10 \mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2} $$ This reaction generates a large amount of heat and gaseous products. It is the sudden formation of these gases, together with their rapid expansion, that produces the explosion. (a) What is the maximum amount of \(\mathrm{O}_{2}\) in grams that can be obtained from \(2.00 \times 10^{2} \mathrm{~g}\) of nitroglycerin? (b) Calculate the percent yield in this reaction if the amount of \(\mathrm{O}_{2}\) generated is found to be \(6.55 \mathrm{~g}\).

Avogadro's number has sometimes been described as a conversion factor between amu and grams. Use the fluorine atom \((19.00\) amu) as an example to show the relationship between the atomic mass unit and the gram.
See all solutions

Recommended explanations on Chemistry Textbooks

Inorganic Chemistry

Read Explanation

Chemical Analysis

Read Explanation

The Earths Atmosphere

Read Explanation

Physical Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation

Organic Chemistry

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.