/*! 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 118 Aluminum metal and iron(III) oxi... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 118

Aluminum metal and iron(III) oxide react to give aluminum oxide and iron metal. What is the maximum mass of iron that can be obtained from a reaction mixture containing \(2.5 \mathrm{g}\) of aluminum and \(9.5 \mathrm{g}\) of iron(III) oxide. What mass of the excess reactant remains?

Short Answer

Expert verified
The maximum mass of iron that can be obtained is 10.39 g. The mass of the excess reactant, iron(III) oxide, that remains is 2.08 g.

Step by step solution

01

Write the balanced chemical equation

Aluminum (Al) reacts with Iron (III) oxide (Fe2O3) to give Aluminum oxide (Al2O3) and Iron (Fe). The balanced chemical equation is: \[2Al + Fe2O3 鈫� Al2O3 + 2Fe\] This equation tells that 2 moles of Al react with 1 mole of Fe2O3 to produce 1 mole of Al2O3 and 2 moles of Fe.
02

Convert mass to moles

Molar mass of Al = 26.98 g/mol. Molar mass of Fe2O3 = 159.7 g/mol. So, Moles of Al = \(\frac{2.5}{26.98} = 0.093\, moles\), Moles of Fe2O3 = \(\frac{9.5}{159.7} = 0.0595\, moles\)
03

Identify the limiting reactant

The reaction requires 2 moles of Al to react with 1 mole of Fe2O3. Based on calculation, there are enough moles of Al for 0.0465 moles of Fe2O3 and enough moles of Fe2O3 for 0.0595 moles of Al. So, Al is the limiting reactant.
04

Determine the amount of iron formed

From the balanced equation, we know that 2 moles of Al form 2 moles of Fe. So, the amount of Fe formed = \(0.093 * 2 = 0.186\, moles\). The molar mass of Fe is 55.85 g/mol, so the mass of Fe formed = \(0.186 * 55.85 = 10.39\, g\)
05

Determine the excess reactant remaining

All the aluminum was used up. For Fe2O3, Initially, we had 0.0595 moles of Fe2O3. The reaction used = \(0.093/2 = 0.0465\, moles\). Remaining Fe2O3 = \(0.0595 - 0.0465 = 0.013\, moles\). Therefore, the mass of Fe2O3 remaining = \(0.013 * 159.7 = 2.08\, g\)

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
Understanding the concept of the limiting reactant is crucial when we're dealing with chemical reactions. Imagine you're trying to make sandwiches鈥攂read slices are like one reactant and cheese slices are like another. If you have 4 slices of bread and only 2 slices of cheese, you can only make 2 sandwiches. The cheese limits the process, and that's what a limiting reactant does in a chemical reaction.

A limiting reactant is the reactant that will be completely used up first, limiting the amount of product that can be formed. To identify it, we first need a balanced chemical equation. Then, by comparing the mole ratio of the reactants used, we can determine which reactant will run out first. Any remaining reactants are known as excess reactants. They're like those extra bread slices waiting in the pantry while the cheese has run out.

In our exercise, after converting the mass of Aluminum (Al) and Iron(III) oxide (Fe2O3) to moles, we discovered that Al would run out first during the reaction. This makes Al the limiting reactant, and thus it dictates the maximum amount of Iron (Fe) that can be produced.
Mole-to-Mass Conversion
Transitioning between moles and mass is a common task in chemistry, much like converting miles to kilometers. This process is known as mole-to-mass conversion and hinges on knowing the substance's molar mass鈥攖he weight of one mole of the substance. To determine the molar mass, we sum the atomic masses of all atoms in the molecule according to its chemical formula.

For the exercise, using the molar masses of Aluminum (26.98 g/mol) and Iron(III) oxide (159.7 g/mol), we convert the given masses into moles. This conversion enables us to use the balanced equation effectively, since it is expressed in moles. From this, we calculate that 2.5 grams of Al yields 0.093 moles, while 9.5 grams of Fe2O3 corresponds to 0.0595 moles. These mole quantities play a pivotal role in finding the limiting reactant and the eventual yield of Iron.
Balanced Chemical Equations
A balanced chemical equation is the Rosetta Stone of chemistry鈥攊t tells us the exact proportion of reactants that combine and the amount of products formed. Think of it as a recipe: for our sandwich, it would be '2 slices of bread + 1 slice of cheese 鈫� 1 sandwich'. Without balance, the equation would be incorrect, and the product could not be formed as expected.

In balancing equations, we ensure that the number of atoms for each element is equal on both the reactant and product sides. This respect for the Law of Conservation of Mass is fundamental; mass can neither be created nor destroyed in a chemical reaction. In the exercise, the balanced equation was \(2Al + Fe2O3 \rightarrow Al2O3 + 2Fe\). From this equation, we see that two moles of Aluminum react with one mole of Iron(III) oxide to form one mole of Aluminum oxide and two moles of Iron, guiding us through the stoichiometric process from reactants to products.

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

For a specific reaction, ammonium dichromate is the only reactant and chromium(III) oxide and water are two of the three products. What is the third product and how many grams of this product are produced per kilogram of ammonium dichromate decomposed?

Explain the important distinctions between (a) chemical formula and chemical equation; (b) stoichiometric coefficient and stoichiometric factor; (c) solute and solvent; (d) actual yield and percent yield; (e) consecutive and simultaneous reactions.

The reaction of calcium hydride with water can be used to prepare small quantities of hydrogen gas, as is done to fill weather-observation balloons. \(\mathrm{CaH}_{2}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \longrightarrow\) $$ \mathrm{Ca}(\mathrm{OH})_{2}(\mathrm{s})+\mathrm{H}_{2}(\mathrm{g}) \text { (not balanced) } $$ (a) How many grams of \(\mathrm{H}_{2}(\mathrm{g})\) result from the reaction of \(127 \mathrm{g} \mathrm{CaH}_{2}\) with an excess of water? (b) How many grams of water are consumed in the reaction of \(56.2 \mathrm{g} \mathrm{CaH}_{2} ?\) (c) What mass of \(\mathrm{CaH}_{2}(\mathrm{s})\) must react with an excess of water to produce \(8.12 \times 10^{24}\) molecules of \(\mathrm{H}_{2} ?\)

High-purity silicon is obtained using a three-step process. The first step involves heating solid silicon dioxide, \(\mathrm{SiO}_{2^{\prime}}\) with solid carbon to give solid silicon and carbon monoxide gas. In the second step, solid silicon is converted into liquid silicon tetrachloride, \(\mathrm{SiCl}_{4}\) by treating it with chlorine gas. In the last step, \(\mathrm{SiCl}_{4}\) is treated with hydrogen gas to give ultrapure solid silicon and hydrogen chloride gas. (a) Write balanced chemical equations for the steps involved in this three- step process. (b) Calculate the masses of carbon, chlorine, and hydrogen required per kilogram of silicon.

The reaction of potassium superoxide, \(\mathrm{KO}_{2}\), is used in life- support systems to replace \(\mathrm{CO}_{2}(\mathrm{g})\) in expired air with \(\mathrm{O}_{2}(\mathrm{g}) .\) The unbalanced chemical equation for the reaction is given below. $$\mathrm{KO}_{2}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g}) \longrightarrow \mathrm{K}_{2} \mathrm{CO}_{3}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{g})$$ (a) How many moles of \(\mathrm{O}_{2}(\mathrm{g})\) are produced by the reaction of \(156 \mathrm{g} \mathrm{CO}_{2}(\mathrm{g})\) with excess \(\mathrm{KO}_{2}(\mathrm{s}) ?\) (b) How many grams of \(\mathrm{KO}_{2}(\mathrm{s})\) are consumed per \(100.0 \mathrm{g} \mathrm{CO}_{2}(\mathrm{g})\) removed from expired air? (c) How many \(\mathrm{O}_{2}\) molecules are produced per milligram of \(\mathrm{KO}_{2}\) consumed?
See all solutions

Recommended explanations on Chemistry Textbooks

Organic Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Making Measurements

Read Explanation

Kinetics

Read Explanation

Chemical Analysis

Read Explanation

Nuclear 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.