Question

Balance each of the following equations that describe decomposition reactions. a. \(\mathrm{NaHCO}_{3}(s) \rightarrow \mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{CO}_{2}(g)\) b. \(\mathrm{NaClO}_{3}(s) \rightarrow \mathrm{NaCl}(s)+\mathrm{O}_{2}(g)\) c. \(\mathrm{HgO}(s) \rightarrow \mathrm{Hg}(l)+\mathrm{O}_{2}(g)\) d. \(C_{12} H_{22} O_{11}(s) \rightarrow C(s)+H_{2} O(g)\) e. \(\mathrm{H}_{2} \mathrm{O}_{2}(l) \rightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g)\)

Short answer

The balanced equations for the given decomposition reactions are as follows: a: \(2 \mathrm{NaHCO}_{3}(s) \rightarrow \mathrm{Na}_{2} \mathrm{CO}_{3}(s) + \mathrm{H}_{2} \mathrm{O}(g) + \mathrm{CO}_{2}(g)\) b: \(2 \mathrm{NaClO}_{3}(s) \rightarrow 2 \mathrm{NaCl}(s) + 3 \mathrm{O}_{2}(g)\) c: \(2 \mathrm{HgO}(s) \rightarrow 2 \mathrm{Hg}(l) + \mathrm{O}_{2}(g)\) d: \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(s) \rightarrow 12 \mathrm{C}(s) + 11 \mathrm{H}_{2} \mathrm{O}(g)\) e: \(2 \mathrm{H}_{2} \mathrm{O}_{2}(l) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) + \mathrm{O}_{2}(g)\)

Step-by-step solution

a. Balancing the first equation: NaHCO₃(s) → Na₂CO₃(s) + H₂O(g) + CO₂(g)

First, let's count the number of atoms of each element present on each side of the equation: Reactants: 1 Na, 2 O, 1 H, 1 C Products: 2 Na, 3 O, 2 H, 2 C Notice that there are different numbers of each atom on both sides of the equation. To balance the equation, we will need to add coefficients step by step. Step 1: Balance Sodium (Na) atoms We can balance Sodium by adding a coefficient of 2 in front of NaHCO₃ on the reactant side. 2 NaHCO₃(s) → Na₂CO₃(s) + H₂O(g) + CO₂(g) Step 2: Balance Oxygen (O) atoms and Carbon (C) atoms Now, the equation is properly balanced, since all the elements have equal numbers of atoms on each side. Reactants: 2 Na, 4 O, 2 H, 2 C Products: 2 Na, 4 O, 2 H, 2 C The balanced equation is: \(2 \mathrm{NaHCO}_{3}(s) \rightarrow \mathrm{Na}_{2} \mathrm{CO}_{3}(s) + \mathrm{H}_{2} \mathrm{O}(g) + \mathrm{CO}_{2}(g)\)

b. Balancing the second equation: NaClO₃(s) → NaCl(s) + O₂(g)

First, let's count the number of atoms of each element present on each side of the equation: Reactants: 1 Na, 3 O, 1 Cl Products: 1 Na, 2 O, 1 Cl Notice that there are different numbers of Oxygen atoms on both sides of the equation. To balance the equation, we will need to add coefficients step by step. Step 1: Balance Oxygen (O) atoms We can balance Oxygen by adding a coefficient of 2 in front of NaClO₃ on the reactant side and a coefficient of 3 in front of O₂ on the product side. 2 NaClO₃(s) → NaCl(s) + 3 O₂(g) Step 2: Balance Sodium (Na) atoms and Chlorine (Cl) atoms Now, the equation is properly balanced, since all the elements have equal numbers of atoms on each side. Reactants: 2 Na, 6 O, 2 Cl Products: 2 Na, 6 O, 2 Cl The balanced equation is: \(2 \mathrm{NaClO}_{3}(s) \rightarrow 2 \mathrm{NaCl}(s) + 3 \mathrm{O}_{2}(g)\)

c. Balancing the third equation: HgO(s) → Hg(l) + O₂(g)

First, let's count the number of atoms of each element present on each side of the equation: Reactants: 1 Hg, 1 O Products: 1 Hg, 2 O Notice that there are different numbers of Oxygen atoms on both sides of the equation. To balance the equation, we will need to add coefficients step by step. Step 1: Balance Oxygen (O) atoms We can balance Oxygen by adding a coefficient of 2 in front of HgO on the reactant side. 2 HgO(s) → Hg(l) + O₂(g) Step 2: Balance Mercury (Hg) atoms Now, we can balance Mercury by adding a coefficient of 2 in front of Hg on the product side. 2 HgO(s) → 2 Hg(l) + O₂(g) Now, the equation is properly balanced, since all the elements have equal numbers of atoms on each side. Reactants: 2 Hg, 2 O Products: 2 Hg, 2 O The balanced equation is: \(2 \mathrm{HgO}(s) \rightarrow 2 \mathrm{Hg}(l) + \mathrm{O}_{2}(g)\)

d. Balancing the fourth equation: C₁₂H₂₂O₁₁(s) → C(s) + H₂O(g)

First, let's count the number of atoms of each element present on each side of the equation: Reactants: 12 C, 22 H, 11 O Products: 1 C, 2 H, 1 O Notice that there are different numbers of each atom on both sides of the equation. To balance the equation, we will need to add coefficients step by step. Step 1: Balance Carbon (C) atoms We can balance Carbon by adding a coefficient of 12 in front of C on the product side. C₁₂H₂₂O₁₁(s) → 12 C(s) + H₂O(g) Step 2: Balance Hydrogen (H) atoms We can balance Hydrogen by adding a coefficient of 11 in front of H₂O on the product side. C₁₂H₂₂O₁₁(s) → 12 C(s) + 11 H₂O(g) Now, the equation is properly balanced, since all the elements have equal numbers of atoms on each side. Reactants: 12 C, 22 H, 11 O Products: 12 C, 22 H, 11 O The balanced equation is: \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(s) \rightarrow 12 \mathrm{C}(s) + 11 \mathrm{H}_{2} \mathrm{O}(g)\)

e. Balancing the fifth equation: H₂O₂(l) → H₂O(l) + O₂(g)

First, let's count the number of atoms of each element present on each side of the equation: Reactants: 2 H, 2 O Products: 2 H, 3 O Notice that there are different numbers of Oxygen atoms on both sides of the equation. To balance the equation, we will need to add coefficients step by step. Step 1: Balance Oxygen (O) atoms We can balance Oxygen by adding a coefficient of 2 in front of H₂O₂ on the reactant side and a coefficient of 2 in front of O₂ on the product side. 2 H₂O₂(l) → H₂O(l) + 2 O₂(g) Step 2: Balance Hydrogen (H) atoms Now, we can balance Hydrogen by adding a coefficient of 2 in front of H₂O on the product side. 2 H₂O₂(l) → 2 H₂O(l) + O₂(g) Now, the equation is properly balanced, since all the elements have equal numbers of atoms on each side. Reactants: 4 H, 4 O Products: 4 H, 4 O The balanced equation is: \(2 \mathrm{H}_{2} \mathrm{O}_{2}(l) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) + \mathrm{O}_{2}(g)\)

Key concepts

Decomposition Reactions

Decomposition reactions are fascinating chemical processes where one compound splits into two or more simpler substances. Understanding these reactions is crucial. This process often requires energy input, such as heat, light, or electricity, to break the bonds in a compound. Decomposition reactions are typically represented by chemical equations where a single reactant yields multiple products. For example, 1. \[NaHCO_3 \rightarrow Na_2CO_3 + H_2O + CO_2\] The above equation is a good illustration of how baking soda (\( NaHCO_3 \)) decomposes to form sodium carbonate, water, and carbon dioxide. Observing how compounds break down can help unravel complex chemical behaviors.In nature, these reactions play a significant role, such as decomposition of organic materials. Be it in classrooms or laboratories, mastering decomposition reactions is a step toward understanding broader chemical principles.

Stoichiometry

Stoichiometry might sound complex, but it's all about measuring and calculating amounts in a chemical equation. It’s the deciphering tool that helps us determine the balanced proportions of reactants and products involved in chemical reactions. The essence of stoichiometry is to ensure that there is the right amount of ingredients to carry out a reaction fully without any leftovers. Let's revisit our balanced equation:\[2 \text{NaHCO}_3 \rightarrow \text{Na}_2\text{CO}_3 + \text{H}_2\text{O} + \text{CO}_2\] Here, stoichiometry dictates that exactly two molecules of baking soda are needed to produce these products. This ensures all the atoms in the reactants are accounted for among the products.To grasp stoichiometry, remember these key steps:

• Count all atoms in the reactants and products.
• Balance the equation so the number of each atom type is equal on both sides.
• Use coefficients to adjust the number of molecules involved.

Understanding stoichiometry can greatly help predict how much of a product can be obtained from given quantities of reactants, making it a cornerstone of chemistry.

Chemical Equations

Chemical equations are the language of chemistry. They represent reactions in a clear, standardized way. Each equation shows the starting reactants and the resulting products, in a balanced manner wherein the mass and charge are conserved. For instance, in the equation \[2 \text{HgO} \rightarrow 2 \text{Hg} + \text{O}_2\] "HgO" represents mercury(II) oxide, "Hg" is mercury, and "O\( _2 \)" is oxygen gas. The arrow signifies the reaction direction, and coefficients before the formulas ensure balance.There are different types of chemical equations, including decomposition, synthesis, and combustion. Decomposition, as we explored earlier, is just one category and involves a single substance breaking down.Chemical equations can be daunting, but following these tips makes them clearer:

• Write each element involved in the reaction.
• Use coefficients to balance the equation, ensuring the same number of each atom on both sides.
• Check your work; accuracy is paramount in chemistry.

Chemical equations offer a concise snapshot of what occurs in a chemical reaction, making them indispensable in science.

Atom Conservation

Atom conservation is a fundamental principle of chemistry, stating that atoms cannot be created or destroyed in a chemical reaction. This is why balancing chemical equations is essential. Every atom that you start with in the reactants must be accounted for in the products.For example, consider the equation:\[C_{12} H_{22} O_{11} \rightarrow 12C + 11H_2O\]Here, you begin with 12 carbon atoms, 22 hydrogen atoms, and 11 oxygen atoms on the reactant side. These must be present in the same amounts on the product side.This principle is crucial for maintaining accuracy and ensuring that nothing is lost "in translation" during a chemical reaction. Points to remember about atom conservation:

• Always count the atoms for each element on both sides of the equation.
• Adjust coefficients as needed to balance the equation.
• Recheck to verify that both sides equal out.

Conserving atoms helps in correctly predicting products, understanding how reactions work, and helps maintain the integrity of chemical equations. It is truly the backbone of the study of chemistry.