Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Write balanced equations for the reactions of aluminum with HCl(aq), \(\mathrm{Cl}_{2},\) and \(\mathrm{O}_{2}\)

Short Answer

Expert verified
Balanced equations: 1. \(2\text{Al} + 6\text{HCl} \rightarrow 2\text{AlCl}_3 + 3\text{H}_2 \) 2. \(2\text{Al} + 3\text{Cl}_2 \rightarrow 2\text{AlCl}_3 \) 3. \(4\text{Al} + 3\text{O}_2 \rightarrow 2\text{Al}_2\text{O}_3 \)

Step by step solution

01

Understand the Reactants and Products

Before writing balanced equations, identify the reactants and products for each reaction. For aluminum reacting with HCl, the products are aluminum chloride and hydrogen gas. For aluminum and chlorine gas, the product is aluminum chloride. For aluminum and oxygen, the product is aluminum oxide.
02

Write Unbalanced Equations

Begin by writing the unbalanced chemical equations for each reaction. 1. Aluminum and HCl: \( \text{Al} + \text{HCl} \rightarrow \text{AlCl}_3 + \text{H}_2 \)2. Aluminum and \( \text{Cl}_2 \): \( \text{Al} + \text{Cl}_2 \rightarrow \text{AlCl}_3 \)3. Aluminum and \( \text{O}_2 \): \( \text{Al} + \text{O}_2 \rightarrow \text{Al}_2\text{O}_3 \)
03

Balance the Equation for Aluminum and HCl

To balance the reaction \( \text{Al} + \text{HCl} \rightarrow \text{AlCl}_3 + \text{H}_2 \), ensure the number of each type of atom is the same on both sides. This yields:\[ 2\text{Al} + 6\text{HCl} \rightarrow 2\text{AlCl}_3 + 3\text{H}_2 \]
04

Balance the Equation for Aluminum and Chlorine Gas

Balance the reaction \( \text{Al} + \text{Cl}_2 \rightarrow \text{AlCl}_3 \) by ensuring an equal number of chlorine atoms. The balanced equation is:\[ 2\text{Al} + 3\text{Cl}_2 \rightarrow 2\text{AlCl}_3 \]
05

Balance the Equation for Aluminum and Oxygen

To balance \( \text{Al} + \text{O}_2 \rightarrow \text{Al}_2\text{O}_3 \), balance the aluminum and oxygen atoms:\[ 4\text{Al} + 3\text{O}_2 \rightarrow 2\text{Al}_2\text{O}_3 \]

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 Vaia!

Key Concepts

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

Chemical Reactions
Chemical reactions involve the transformation of reactants into products through the rearrangement of atoms. When aluminum reacts with various substances, the bonds between atoms break and new bonds form, leading to the formation of entirely new compounds. In our exercise, aluminum reacts with hydrochloric acid, chlorine gas, and oxygen, demonstrating different types of chemical reactions. These can include synthesis and oxidation reactions, showcasing the versatility of aluminum as a reactant. Each reaction involves:
  • Identifying the reactants and the expected products.
  • Writing the chemical formulae correctly for each compound involved.
  • Understanding what changes are taking place at the atomic level.
Observe how reactants and products might differ significantly in physical and chemical properties after the reaction.
Stoichiometry
Stoichiometry is the study of quantitative relationships in chemical reactions. It allows us to calculate how much of each reactant is needed and how much of each product is formed. In this exercise, stoichiometry helps us understand the proportions of aluminum and other reactants like HCl, chlorine, and oxygen. It ensures that when these substances react, they do so in a balanced manner so that the laws of conservation of mass are respected. Key steps include:
  • Identifying the mole ratio between reactants and products.
  • Using conversion factors to relate masses of reactants to moles, and further to molecules.
  • Calculating the required amount of each reactant to completely react without any leftover substance.
These calculations are vital in predicting the outcomes of reactions and ensuring efficiency in chemical processes.
Aluminum Reactions
Aluminum reacts with various elements and compounds, displaying its chemical reactivity. It is a lightweight and strong metal, widely used in many industries, which also makes it a common subject in chemistry exercises. When reacting with hydrochloric acid, aluminum turns into aluminum chloride and releases hydrogen gas, a classic displacement reaction. In the case of chlorine gas, aluminum forms aluminum chloride through a combination reaction where the elements directly combine. Lastly, aluminum reacting with oxygen forms aluminum oxide, a protective layer often utilized to prevent further corrosion. Such reactions show:
  • The diverse nature of aluminum's chemical behavior.
  • Its role in forming both ionic and covalent compounds.
  • The formation of protective scales of aluminum oxide that safeguard the metal surface from corrosion.
Aluminum's reactions not only show its potential uses but also its protective properties in various applications.
Chemical Equation Balancing
Balancing chemical equations is a crucial skill in chemistry, reflecting the conservation of mass. By balancing equations, we ensure that the number of each type of atom is the same on both sides of the reaction. This practice is essential for accurately describing chemical reactions. In the provided solutions, balancing was done as follows:
  • Balancing the equation for aluminum reacting with HCl was achieved by ensuring the same number of Al, Cl, and H atoms on both sides.
  • The equation with aluminum and chlorine required balancing the chlorine atoms first before verifying aluminum.
  • Aluminum's reaction with oxygen involved adjusting both Al and O atoms, achieving an equal balance on both sides of the equation.
A properly balanced equation ensures correct stoichiometric calculations and understanding of the reaction dynamics. This is pivotal in fields ranging from industrial chemistry to environmental science.

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

Worldwide production of silicon carbide, \(\mathrm{SiC},\) is several hundred thousand tons annually. If you want to produce \(1.0 \times 10^{5}\) metric tons of \(\mathrm{SiC}\) what mass (metric tons) of silicon sand \(\left(\mathrm{SiO}_{2}\right)\) will you use if \(70 \%\) of the sand is converted to SiC? (A metric ton is exactly \(1000 \mathrm{kg} .\) )

Compare the mass of \(\mathrm{H}_{2}\) expected from the reaction of steam \(\left(\mathrm{H}_{2} \mathrm{O}\right)\) per mole of methane, petroleum, and coal. (Assume complete reaction in each case. Use \(\mathrm{CH}_{2}\) and \(\mathrm{CH}\) as representative formulas for petroleum and coal, respectively.

The reaction of elemental phosphorus and excess oxygen produces \(\mathrm{P}_{4} \mathrm{O}_{10} .\) Name the compound. (a) phosphorus oxide (b) phosphoric acid (c) phosphorus decaoxide (d) tetraphosphorus decaoxide

A method recently suggested for the preparation of hydrogen (and oxygen) from water proceeds as follows: (a) Sulfuric acid and hydrogen iodide are formed from sulfur dioxide, water, and iodine. (b) The sulfuric acid from the first step is decomposed by heat to water, sulfur dioxide, and oxygen. (c) The hydrogen iodide from the first step is decomposed with heat to hydrogen and iodine. Write a balanced equation for each of these steps, and show that their sum is the decomposition of water to form hydrogen and oxygen.

Complete and balance the equations for the following reactions. [Assume an excess of oxygen for (d).] (a) \(\mathrm{Na}(\mathrm{s})+\mathrm{Br}_{2}(\ell) \rightarrow\) (c) \(\mathrm{Al}(\mathrm{s})+\mathrm{F}_{2}(\mathrm{g}) \rightarrow\) (b) \(\mathrm{Mg}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{g}) \rightarrow\) (d) \(\mathrm{C}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{g}) \rightarrow\)

See all solutions

Recommended explanations on Chemistry Textbooks

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