Question

Balance each of the following chemical equations. a. \(\mathrm{KO}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{KOH}(a q)+\mathrm{O}_{2}(g)+\mathrm{H}_{2} \mathrm{O}_{2}(a q)\) b. \(\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+\mathrm{HNO}_{3}(a q) \rightarrow \mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\) c. \(\mathrm{NH}_{3}(g)+\mathrm{O}_{2}(g) \rightarrow \mathrm{NO}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) d. \(\mathrm{PCl}_{5}(l)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{H}_{3} \mathrm{PO}_{4}(a q)+\mathrm{HCl}(g)\) e. \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)\) f. \(\mathrm{CaO}(s)+\mathrm{C}(s) \rightarrow \mathrm{CaC}_{2}(s)+\mathrm{CO}_{2}(g)\) g. \(\operatorname{MoS}_{2}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{MoO}_{3}(s)+\mathrm{SO}_{2}(g)\) h. \(\mathrm{FeCO}_{3}(s)+\mathrm{H}_{2} \mathrm{CO}_{3}(a q) \rightarrow \mathrm{Fe}\left(\mathrm{HCO}_{3}\right)_{2}(a q)\)

Short answer

The balanced chemical equations are as follows: a. \(\mathrm{KO}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow 2\mathrm{KOH}(a q)+\mathrm{O}_{2}(g)+\mathrm{H}_{2} \mathrm{O}_{2}(a q)\) b. \(\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+6\mathrm{HNO}_{3}(a q) \rightarrow \mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3}(a q)+3\mathrm{H}_{2} \mathrm{O}(l)\) c. \(4\mathrm{NH}_{3}(g)+5\mathrm{O}_{2}(g) \rightarrow 4\mathrm{NO}(g)+6\mathrm{H}_{2} \mathrm{O}(g)\) d. \(\mathrm{PCl}_{5}(l)+4\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{H}_{3} \mathrm{PO}_{4}(a q)+5\mathrm{HCl}(g)\) e. \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+3\mathrm{O}_{2}(g) \rightarrow 2\mathrm{CO}_{2}(g)+3\mathrm{H}_{2} \mathrm{O}(l)\) f. \(\mathrm{CaO}(s)+3\mathrm{C}(s) \rightarrow \mathrm{CaC}_{2}(s)+\mathrm{CO}_{2}(g)\) g. \(2\operatorname{MoS}_{2}(s)+7\mathrm{O}_{2}(g) \rightarrow 2\mathrm{MoO}_{3}(s)+4\mathrm{SO}_{2}(g)\) h. \(\mathrm{FeCO}_{3}(s)+2\mathrm{H}_{2} \mathrm{CO}_{3}(a q) \rightarrow \mathrm{Fe}\left(\mathrm{HCO}_{3}\right)_{2}(a q)\)

Step-by-step solution

Count the atoms on both sides

On the reactant side: K=1, O=3, H=2 On the product side: K=1, O=4, H=3

Adjust coefficients

Add a coefficient of 2 in front of KOH on the product side: \(\mathrm{KO}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow 2\mathrm{KOH}(a q)+\mathrm{O}_{2}(g)+\mathrm{H}_{2} \mathrm{O}_{2}(a q)\) Now we have: K=1*2=2, O=2+2=4, H=2. This equation is now balanced. b. Balancing \(\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+\mathrm{HNO}_{3}(a q) \rightarrow \mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\):

Count the atoms on both sides

On the reactant side: Fe=2, O=4, H=1, N=1 On the product side: Fe=1, O=12, H=2, N=3

Adjust coefficients

Add coefficients of 6 in front of HNO3 on the reactant side, and 3 in front of H2O on the product side: \(\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+6\mathrm{HNO}_{3}(a q) \rightarrow \mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3}(a q)+3\mathrm{H}_{2} \mathrm{O}(l)\) Now we have: Fe=2, O=3+18=21, H=1*6=6, N=1*6=6. This equation is now balanced. [... Balancing remaining equations the same way ...] c. Balanced equation: \(4\mathrm{NH}_{3}(g)+5\mathrm{O}_{2}(g) \rightarrow 4\mathrm{NO}(g)+6\mathrm{H}_{2} \mathrm{O}(g)\) d. Balanced equation: \(\mathrm{PCl}_{5}(l)+4\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{H}_{3} \mathrm{PO}_{4}(a q)+5\mathrm{HCl}(g)\) e. Balanced equation: \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+3\mathrm{O}_{2}(g) \rightarrow 2\mathrm{CO}_{2}(g)+3\mathrm{H}_{2} \mathrm{O}(l)\) f. Balanced equation: \(\mathrm{CaO}(s)+3\mathrm{C}(s) \rightarrow \mathrm{CaC}_{2}(s)+\mathrm{CO}_{2}(g)\) g. Balanced equation: \(2\operatorname{MoS}_{2}(s)+7\mathrm{O}_{2}(g) \rightarrow 2\mathrm{MoO}_{3}(s)+4\mathrm{SO}_{2}(g)\) h. Balanced equation: \(\mathrm{FeCO}_{3}(s)+2\mathrm{H}_{2} \mathrm{CO}_{3}(a q) \rightarrow \mathrm{Fe}\left(\mathrm{HCO}_{3}\right)_{2}(a q)\)

Key concepts

Stoichiometry

Stoichiometry is a fundamental concept in chemistry that involves calculating the quantities of reactants and products in chemical reactions. Imagine it like a recipe where the ingredients must be combined in precise amounts to get the desired dish. Similarly, in chemical reactions, reactants must be measured in accurate proportions to yield products.

Key aspects of stoichiometry include:

• Balanced Equations: To use stoichiometry, the chemical equation must first be balanced. This means the number of each type of atom is the same on both sides of the equation.
• Mole Ratios: Once balanced, the equation provides the mole ratios of reactants and products, crucial for calculations. For example, in the combustion of methane \(\mathrm{CH}_{4} + 2 \mathrm{O}_{2} \rightarrow \mathrm{CO}_{2} + 2 \mathrm{H}_{2}\mathrm{O}\), the ratio is critical for determining how much oxygen is needed for a given amount of methane.
• Conversions: Stoichiometry often involves converting between units like grams, moles, and molecules using the molar mass of substances.

Understanding stoichiometry is essential for predicting product amounts, optimizing reactions, and scaling up processes from laboratory to industrial scales.

Law of Conservation of Mass

The Law of Conservation of Mass is an essential principle in chemistry, stating that mass is neither created nor destroyed in a chemical reaction. Instead, it is conserved. This means that the total mass of reactants before the reaction must equal the total mass of products after the reaction.

An example of this is observed in the first balanced equation from the exercise: \(\mathrm{KO}_{2}(s)+\mathrm{H}_{2}\mathrm{O}(l) \rightarrow 2\mathrm{KOH}(aq)+\mathrm{O}_{2}(g)+\mathrm{H}_{2}\mathrm{O}_{2}(aq)\).In this reaction, the total number of each type of atom on the reactant side matches the product side.

Here are some key points:

• Unchanged Matter: The matter may change form, such as solid to gas, but the total mass stays the same.
• Equation Balancing: Ensures the law is followed by making sure the number of atoms for each element is equal on both sides of the reaction.
• Practical Implications: Crucial for predicting how much of a reactant is needed or how much product will be made.

This law underlies all chemical equations and ensures that no atoms are lost during reactions, allowing precise predictions of amounts needed and produced.

Chemical Reactions

Chemical reactions involve the transformation of substances through breaking and forming of bonds, resulting in new products. Each chemical reaction is represented by a chemical equation which shows the reactants and products along with their physical states.

Some important features of chemical reactions include:

• Reactants and Products: Reactants are substances that start the reaction, and products are what we get at the end. In the candle combustion reaction, wax (a reactant) turns into carbon dioxide and water (products).
• Energy Changes: Reactions can either release energy (exothermic) or absorb energy (endothermic). Understanding these changes is vital for applications like energy production.
• Reaction Types: Common reaction types include synthesis, decomposition, single replacement, and double replacement. Each type has its own characteristics and patterns.

Chemical reactions are at the heart of all chemical processes, enabling everything from breathing to industrial manufacturing. Understanding how they occur and are represented through equations is fundamental for any aspiring chemist.