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A common lecture demonstration called "elephant's toothpaste" demonstrates the reaction of hydrogen peroxide producing water and oxygen gas. Write the unbalanced chemical equation for this process.

Short Answer

Expert verified
The unbalanced chemical equation for the decomposition of hydrogen peroxide into water and oxygen gas is \(H_2O_2 (l) \rightarrow H_2O (l) + O_2 (g)\).

Step by step solution

01

Identify the reactants and products

The given reaction involves the decomposition of hydrogen peroxide (H₂O₂) into water (H₂O) and oxygen gas (O₂). Step 2: Write the unbalanced chemical equation
02

Write the unbalanced chemical equation

Now that we know the reactants and products involved, we can write the unbalanced chemical equation for the decomposition of hydrogen peroxide: \(H_2O_2 (l) \rightarrow H_2O (l) + O_2 (g)\) In this equation, "l" denotes the liquid state and "g" denotes the gaseous state. The unbalanced equation represents the reactants and products, but the numbers of atoms on both sides are not equal, so the equation is not yet balanced.

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Key Concepts

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

Decomposition Reactions
Decomposition reactions are a type of chemical reaction in which a single compound breaks down into two or more simpler substances. These reactions are characterized by the general formula:
\( AB \rightarrow A + B \)
where \( AB \) represents the compound that undergoes decomposition, and \( A \) and \( B \) are the products formed. In the context of the 'elephant's toothpaste' demonstration, the compound hydrogen peroxide (\( H_2O_2 \)) decomposes into water (\( H_2O \)) and oxygen gas (\( O_2 \)), showcasing a classic example of a decomposition reaction.

Decomposition reactions can be triggered by various stimuli, such as heat, light, or the presence of a catalyst. In 'elephant's toothpaste', a catalyst is typically used to speed up the reaction, making it more visually dramatic. This type of reaction is important in many industrial and biological processes. For example, the decomposition of organic matter is essential for nutrient cycling in ecosystems.
Balancing Chemical Equations
Balancing chemical equations is a fundamental skill in chemistry that ensures the Law of Conservation of Mass is upheld. This law states that in a closed system, mass cannot be created or destroyed. Therefore, for a chemical equation to accurately represent a chemical reaction, the number of atoms for each element must be the same on both sides of the equation.

To balance an equation, one must adjust the coefficients, which are numbers placed in front of compounds or elements, in order to make the number of atoms of each element equal on both sides. It is essential not to alter the chemical formulas of the reactants or products while balancing the equation. Continuing with our example, the unbalanced equation for the decomposition of hydrogen peroxide is:
\( H_2O_2 (l) \rightarrow H_2O (l) + O_2 (g) \)
To balance this, we would need to ensure that there are equal numbers of hydrogen and oxygen atoms on both sides. One possible balanced equation is:
\( 2H_2O_2 (l) \rightarrow 2H_2O (l) + O_2 (g) \)
Now, there are four hydrogen atoms and four oxygen atoms on each side of the equation, maintaining mass balance. Balancing equations is crucial for quantitatively describing the reactants consumed and the products formed in a chemical reaction.
Reactants and Products
Reactants and products are the substances involved at the beginning and end of a chemical reaction, respectively. Reactants are the substances that undergo a chemical change, and the products are the new substances formed as a result of that change.

In a chemical equation, reactants are typically written on the left side, and the products are written on the right side, with an arrow pointing from reactants to products symbolizing the direction of the reaction. In the 'elephant's toothpaste' demonstration, hydrogen peroxide (\( H_2O_2 \)) serves as the reactant that decomposes, while water (\( H_2O \)) and oxygen gas (\( O_2 \)) are the products. Understanding the roles of reactants and products is essential for analyzing and predicting the outcomes of chemical reactions. It also forms the basis for stoichiometry, which involves the calculation of the relative quantities of reactants and products involved in a reaction.

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Most popular questions from this chapter

When hydrogen sulfide, \(\mathrm{H}_{2} \mathrm{~S}\), gas is bubbled through a solution of lead(II) nitrate, \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\), a black precipitate of lead(II) sulfide, \(\mathrm{PbS},\) forms, and nitric acid, \(\mathrm{HNO}_{3},\) is produced. Write the unbalanced chemical equation for this reaction.

Iron oxide ores, commonly a mixture of \(\mathrm{FeO}\) and \(\mathrm{Fe}_{2} \mathrm{O}_{3}\), are given the general formula \(\mathrm{Fe}_{3} \mathrm{O}_{4}\). They yield elemental iron when heated to a very high temperature with either carbon monoxide or elemental hydrogen. Balance the following equations for these processes. $$\begin{aligned}\mathrm{Fe}_{3} \mathrm{O}_{4}(s)+\mathrm{H}_{2}(g) & \rightarrow \mathrm{Fe}(s)+\mathrm{H}_{2} \mathrm{O}(g) \\ \mathrm{Fe}_{3} \mathrm{O}_{4}(s)+\mathrm{CO}(g) & \rightarrow \mathrm{Fe}(s)+\mathrm{CO}_{2}(g)\end{aligned}$$

Pure silicon, which is needed in the manufacturing of electronic components, may be prepared by heating silicon dioxide (sand) with carbon at high temperatures, releasing carbon monoxide gas. Write the unbalanced chemical equation for this process.

Balance each of the following chemical equations. a. \(\mathrm{SiCl}_{4}(l)+\mathrm{Mg}(s) \rightarrow \mathrm{Si}(s)+\mathrm{MgCl}_{2}(s)\) b. \(\mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \rightarrow \mathrm{NOCl}(g)\) c. \(\operatorname{MnO}_{2}(s)+\operatorname{Al}(s) \rightarrow \operatorname{Mn}(s)+\mathrm{Al}_{2} \mathrm{O}_{3}(s)\) d. \(\operatorname{Cr}(s)+\mathrm{S}_{8}(s) \rightarrow \mathrm{Cr}_{2} \mathrm{~S}_{3}(s)\) e. \(\mathrm{NH}_{3}(g)+\mathrm{F}_{2}(g) \rightarrow \mathrm{NH}_{4} \mathrm{~F}(s)+\mathrm{NF}_{3}(g)\) f. \(\mathrm{Ag}_{2} \mathrm{~S}(s)+\mathrm{H}_{2}(g) \rightarrow \mathrm{Ag}(s)+\mathrm{H}_{2} \mathrm{~S}(g)\) g. \(\mathrm{O}_{2}(g) \rightarrow \mathrm{O}_{3}(g)\) h. \(\mathrm{Na}_{2} \mathrm{SO}_{3}(a q)+\mathrm{S}_{8}(s) \rightarrow \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}(a q)\)

Balance each of the following chemical equations. a. \(\mathrm{Fe}_{3} \mathrm{O}_{4}(s)+\mathrm{H}_{2}(g) \rightarrow \mathrm{Fe}(l)+\mathrm{H}_{2} \mathrm{O}(g)\) b. \(\mathrm{K}_{2} \mathrm{SO}_{4}(a q)+\mathrm{BaCl}_{2}(a q) \rightarrow \mathrm{BaSO}_{4}(s)+\mathrm{KCl}(a q)\) c. \(\mathrm{HCl}(a q)+\mathrm{FeS}(s) \rightarrow \mathrm{FeCl}_{2}(a q)+\mathrm{H}_{2} \mathrm{~S}(g)\) d. \(\mathrm{Br}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{SO}_{2}(g) \rightarrow \mathrm{HBr}(a q)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q)\) e. \(\mathrm{CS}_{2}(l)+\mathrm{Cl}_{2}(g) \rightarrow \mathrm{CCl}_{4}(l)+\mathrm{S}_{2} \mathrm{Cl}_{2}(g)\) f. \(\mathrm{Cl}_{2} \mathrm{O}_{7}(g)+\mathrm{Ca}(\mathrm{OH})_{2}(a q) \rightarrow \mathrm{Ca}\left(\mathrm{ClO}_{4}\right)_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\) g. \(\operatorname{PBr}_{3}(l)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{H}_{3} \mathrm{PO}_{3}(a q)+\mathrm{HBr}(g)\) h. \(\mathrm{Ba}\left(\mathrm{ClO}_{3}\right)_{2}(s) \rightarrow \mathrm{BaCl}_{2}(s)+\mathrm{O}_{2}(s)\)

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