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Write chemical equations to represent the most probable outcome in each of the following. If no reaction is likely to occur, so state. (a) \(\mathrm{B}(\mathrm{OH})_{3} \stackrel{\Delta}{\longrightarrow}\) (b) \(\mathrm{Al}_{2} \mathrm{O}_{3}(\mathrm{s}) \stackrel{\Delta}{\longrightarrow}\) (c) \(\mathrm{CaSO}_{4} \cdot 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{s}) \stackrel{\Delta}{\longrightarrow}\)

Short Answer

Expert verified
(a) \(\mathrm{B(OH)_3}\) \(\stackrel{\Delta}{\longrightarrow}\) \(\mathrm{B_2O_3}\) + \(3 \mathrm{H_2O}\) (b) \(\mathrm{Al_2O_3}\) \(\stackrel{\Delta}{\longrightarrow}\) no reaction (c) \(\mathrm{CaSO_4 \cdot 2 H_2O}\) \(\stackrel{\Delta}{\longrightarrow} \mathrm{CaSO_4 \cdot 0.5 H_2O}\) + \(1.5 \mathrm{H_2O}\)

Step by step solution

01

Analyze boron hydroxide

Heat causes boron hydroxide (\(\mathrm{B(OH)_3}\)) to decompose into boron trioxide and water. The chemical equation is: \(\mathrm{B(OH)_3}\) \( \stackrel{\Delta}{\longrightarrow}\) \(\mathrm{B_2O_3}\) + \(3 \mathrm{H_2O}\)
02

Analyze aluminum oxide

Aluminium oxide (\(\mathrm{Al_2O_3}\)) is a very stable compound and does not decompose upon heating. Thus, no reaction occurs, which we can write as: \(\mathrm{Al_2O_3}\) \(\stackrel{\Delta}{\longrightarrow}\) no reaction
03

Analyze Calcium sulfate dihydrate

Calcium sulfate dihydrate (\(\mathrm{CaSO_4 \cdot 2 H_2O}\)) loses two water molecules upon heating to form Calcium sulfate hemihydrate, the equation being: \(\mathrm{CaSO_4 \cdot 2 H_2O}\) \(\stackrel{\Delta}{\longrightarrow} \mathrm{CaSO_4 \cdot 0.5 H_2O}\) + \(1.5 \mathrm{H_2O}\)

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Key Concepts

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

Decomposition Reactions
Decomposition reactions play a crucial role in understanding how compounds break down. When a compound undergoes a decomposition reaction, it splits into simpler substances or compounds. This type of chemical reaction often requires an input of energy such as heat or light to occur.

Imagine you have a complex Lego structure; when you take it apart, the individual Lego pieces represent the simpler substances resulting from decomposition.

Typically, decomposition reactions can be identified when a single compound breaks down into two or more products. Let's take a look at the types of decomposition reactions you might encounter:
  • Thermal decomposition: Requires heat to break down the compound, like heating boron hydroxide to produce boron trioxide and water.
  • Electrolytic decomposition: Uses electricity to decompose a compound, such as the electrolysis of water into hydrogen and oxygen gases.
  • Photolytic decomposition: Involves the use of light, often seen in the breakdown of silver bromide in photographic films.
Understanding decomposition reactions can help you predict what happens when a compound is heated or exposed to other energy forms.
Thermal Decomposition
Thermal decomposition is a specific type of decomposition reaction driven by the addition of heat. This heat provides the energy necessary to break chemical bonds in a compound, leading it to decompose into simpler substances.

For example, heating boron hydroxide ( B(OH)_3 ) causes it to decompose into boron trioxide ( B_2O_3 ) and water ( H_2O ). This reaction occurs because the supplied heat energy helps weaken and eventually break bonds within the compound, facilitating the formation of new, smaller molecules.

Not all compounds decompose upon heating. A compound like aluminium oxide ( Al_2O_3 ) remains stable even when exposed to high temperatures. This stability shows that the energy supplied might not be sufficient to overcome the strong bonds within its structure.

Thermal decomposition is quite significant in various industrial processes such as refining metals, producing ceramics, and creating cement. Understanding this reaction helps you appreciate the transformations happening at a molecular level when heat is applied.
Chemical Equations
Chemical equations provide a concise way to express what happens during a chemical reaction. In a chemical equation, the substances on the left are the reactants, and those on the right are the products. An arrow (\rightarrow) separates them, indicating the direction of the reaction.

For instance, the chemical equation for the thermal decomposition of calcium sulfate dihydrate is:\[\mathrm{CaSO_4 \cdot 2H_2O} \stackrel{\Delta}{\longrightarrow} \mathrm{CaSO_4 \cdot 0.5H_2O} + 1.5 \mathrm{H_2O}\]This equation tells us that heating calcium sulfate dihydrate turns it into calcium sulfate hemihydrate and water.

Balancing chemical equations is essential to reflect the law of conservation of mass. This law states that mass is neither created nor destroyed in a chemical reaction, meaning the number of atoms of each element should be the same on both sides of the equation. Understanding how to write and balance equations helps you accurately depict chemical processes, predict the products of reactions, and understand stoichiometric relationships in the reaction.

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Most popular questions from this chapter

The first electrolytic process to produce sodium metal used molten NaOH as the electrolyte. Write probable half-equations and an overall equation for this electrolysis.

Describe and explain the similarities and differences between the reaction of a silicate with an acid and that of a carbonate with an acid.

Listed are several pairs of substances. For some pairs, one or both members of the pair react individually with water to produce a gas. For others, neither member of the pair reacts with water. The pair for which each member reacts with water and yields the same gaseous product is (a) \(\mathrm{Al}(\mathrm{s})\) and \(\mathrm{Ba}(\mathrm{s}) ;\) (b) \(\mathrm{Ca}(\mathrm{s})\) and \(\mathrm{CaH}_{2}(\mathrm{s}) ; \quad(\mathrm{c}) \quad \mathrm{Na}(\mathrm{s})\) and \(\mathrm{Na}_{2} \mathrm{O}_{2}(\mathrm{s}) ; \quad\) (d) \(\mathrm{K}(\mathrm{s})\) and \(\mathrm{KO}_{2}(\mathrm{s}) ;(\mathrm{e}) \mathrm{NaHCO}_{3}(\mathrm{s})\) and \(\mathrm{HCl}(\mathrm{aq}).\)

An analysis of a Solvay-process plant shows that for every \(1.00 \mathrm{kg}\) of \(\mathrm{NaCl}\) consumed, \(1.03 \mathrm{kg}\) of \(\mathrm{NaHCO}_{3}\) are obtained. The quantity of \(\mathrm{NH}_{3}\) consumed in the overall process is \(1.5 \mathrm{kg}.\) (a) What is the percent efficiency of this process for converting NaCl to \(\mathrm{NaHCO}_{3} ?\) (b) Why is so little \(\mathrm{NH}_{3}\) required?

Show that, in principle, \(\mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{aq})\) can be converted almost completely to NaOH(aq) by the reaction \(\mathrm{Ca}(\mathrm{OH})_{2}(\mathrm{s})+\mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{aq}) \longrightarrow\) $$ \mathrm{CaCO}_{3}(\mathrm{s})+2 \mathrm{NaOH}(\mathrm{aq}) $$

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