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Write a balanced equation for the reaction of aluminum hydroxide with sulfuric acid.

Short Answer

Expert verified
The balanced equation is \( 2\text{Al(OH)}_3 + 3\text{H}_2\text{SO}_4 \rightarrow \text{Al}_2(\text{SO}_4)_3 + 6\text{H}_2\text{O} \).

Step by step solution

01

Write the Formula for Each Reactant

Start by determining the chemical formula for each reactant. Aluminum hydroxide is represented by \( \text{Al(OH)}_3 \), and sulfuric acid is represented by \( \text{H}_2\text{SO}_4 \).
02

Predict the Products of the Reaction

Aluminum hydroxide reacts with sulfuric acid to form aluminum sulfate and water. The formula for aluminum sulfate is \( \text{Al}_2(\text{SO}_4)_3 \). Water is represented as \( \text{H}_2\text{O} \).
03

Write the Unbalanced Equation

Write the equation with the reactants and products: \[ \text{Al(OH)}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{Al}_2(\text{SO}_4)_3 + \text{H}_2\text{O} \]
04

Balance the Equation

Start by balancing aluminum. We have two aluminums on the right, so put a coefficient of 2 in front of \( \text{Al(OH)}_3 \). Now, balance the sulfate ions. We need three \( \text{H}_2\text{SO}_4 \) molecules to have three sulfate ions for \( \text{Al}_2(\text{SO}_4)_3 \). Lastly, balance water. The entire balanced equation is:\[ 2\text{Al(OH)}_3 + 3\text{H}_2\text{SO}_4 \rightarrow \text{Al}_2(\text{SO}_4)_3 + 6\text{H}_2\text{O} \]

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Key Concepts

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

Understanding Aluminum Hydroxide
Aluminum hydroxide, denoted as \( \text{Al(OH)}_3 \), is a chemical compound consisting of aluminum and hydroxide groups. This compound is often found as a white, gelatinous substance. Aluminum hydroxide is widely used in various industrial processes and everyday products. It's commonly used as an antacid to neutralize stomach acid. Another significant use is in water purification, acting as a flocculating agent.

In chemical reactions, aluminum hydroxide often acts as a base because it can donate hydroxide ions \( (OH^- ) \). When it reacts with acids, such as sulfuric acid, it undergoes neutralization, forming water and other compounds. This reaction releases Aluminum ions which can then form new compounds with other available ions.
Reaction with Sulfuric Acid
Sulfuric acid, with the chemical formula \( \text{H}_2\text{SO}_4 \), is a strong acid known for its ability to react with a wide variety of substances. When sulfuric acid reacts with aluminum hydroxide, it leads to the formation of aluminum sulfate and water. This type of reaction is known as a neutralization reaction, where an acid and a base react to form water and a salt. In the context of aluminum hydroxide and sulfuric acid:
  • Aluminum hydroxide \( (\text{Al(OH)}_3) \) acts as the base.
  • Sulfuric acid \( (\text{H}_2\text{SO}_4) \) acts as the acid.
  • Aluminum sulfate \( (\text{Al}_2(\text{SO}_4)_3) \) is the salt formed.
  • Water \( \text{H}_2\text{O} \) is the result of the neutralization process.
This reaction is a classic illustration of acid-base chemistry in action.
Introducing Stoichiometry
Stoichiometry is a fundamental concept in chemistry that involves the calculation of reactants and products in chemical reactions. It provides the quantitative relationship between the amounts of reactants used and products formed. Understanding stoichiometry is essential when balancing chemical equations to ensure that the number of atoms for each element is consistent on both sides of the equation.

In the reaction between aluminum hydroxide and sulfuric acid, stoichiometry requires us to:
  • Count how many aluminum \( (\text{Al}) \) atoms are present.
  • Ensure equal numbers of sulfate \( (\text{SO}_4^{2-}) \) ions.
  • Balance the number of water \( (\text{H}_2\text{O}) \) molecules produced.
By applying stoichiometry, we ensure the chemical equation accurately reflects the conservation of mass, meaning no atoms are lost or gained during the reaction.
Key Elements of Chemical Reactions
Chemical reactions involve changes where substances, known as reactants, transform into new substances, called products. These transformations adhere to the laws of chemistry, particularly the law of conservation of mass, stating that mass is neither created nor destroyed in a chemical reaction. Key features include:
  • Reactants: Initial substances that undergo change. In our context, they are aluminum hydroxide and sulfuric acid.
  • Products: New substances formed, which are aluminum sulfate and water in this case.
  • Reaction Conditions: Factors such as temperature, pressure, or presence of catalysts can affect the speed and outcome of reactions.
Balancing equations is crucial as it shows how reactants combine to form products, ensuring that the same number of each type of atom appears on both sides of the equation. This balance is vital for correctly understanding the reaction's stoichiometry and practical applications in real-world uses.

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Most popular questions from this chapter

Predict the products of each precipitation reaction. Balance the completed equation, and then write the net ionic equation. (a) \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})+\mathrm{KBr}(\mathrm{aq}) \rightarrow\) (b) \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})+\mathrm{KF}(\mathrm{aq}) \rightarrow\) (c) \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})+\mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(\mathrm{aq}) \rightarrow\)

Balance the following equations, and then classify each as a precipitation, acid-base, or gas-forming reaction. Show states for the products (s, \(\ell, \mathrm{g},\) aq), and then balance the completed equation. Write the net ionic equation. (a) \(\mathrm{Fe}(\mathrm{OH})_{3}(\mathrm{s})+\mathrm{HNO}_{3}(\mathrm{aq}) \rightarrow \mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3}+\mathrm{H}_{2} \mathrm{O}\) (b) \(\mathrm{FeCO}_{3}(\mathrm{s})+\mathrm{HNO}_{3}(\mathrm{aq}) \rightarrow \mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{2}+\mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O}\)

The following reaction can be used to prepare iodine in the laboratory. $$\begin{aligned}2 \mathrm{NaI}(\mathrm{s}) &+2 \mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})+\mathrm{MnO}_{2}(\mathrm{s}) \rightarrow \\\& \mathrm{Na}_{2} \mathrm{SO}_{4}(\mathrm{aq})+\mathrm{MnSO}_{4}(\mathrm{aq})+\mathrm{I}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\ell)\end{aligned}$$ (a) Determine the oxidation number of each atom in the equation. (b) What is the oxidizing agent, and what has been oxidized? What is the reducing agent, and what has been reduced? (c) Is the reaction observed product-favored or reactant-favored? (d) Name the reactants and products.

Identify, from each list below, the compound or compounds that will dissolve in water to give a solution that is only a very weak conductor of electricity. (a) \(\mathrm{NH}_{3}, \mathrm{NaOH}, \mathrm{Ba}(\mathrm{OH})_{2}, \mathrm{Fe}(\mathrm{OH})_{3}\) (b) \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}, \mathrm{Na}_{3} \mathrm{PO}_{4}, \mathrm{HF}, \mathrm{HNO}_{3}\)

Determine the oxidation number of each element in the following ions or compounds. (a) \(\mathrm{PF}_{6}^{-}\) (b) \(\mathrm{H}_{2} \mathrm{AsO}_{4}^{-}\) (c) \(\mathrm{UO}^{2+}\) (d) \(\mathrm{N}_{2} \mathrm{O}_{5}\) (e) \(\mathrm{POCl}_{3}\) (f) \(\mathrm{XeO}_{4}^{2-}\)

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