Question

Balance each of the following chemical equations. a. \(\mathrm{FeCl}_{3}(a q)+\mathrm{KOH}(a q) \rightarrow \mathrm{Fe}(\mathrm{OH})_{3}(s)+\mathrm{KCl}(a q)\) b. \(\mathrm{Pb}\left(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)_{2}(a q)+\mathrm{KI}(a q) \rightarrow \mathrm{PbI}_{2}(s)+\mathrm{KC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(a q)\) c. \(\mathrm{P}_{4} \mathrm{O}_{10}(s)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{H}_{3} \mathrm{PO}_{4}(a q)\) d. \(\mathrm{Li}_{2} \mathrm{O}(s)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{LiOH}(a q)\) e. \(\operatorname{MnO}_{2}(s)+\mathrm{C}(s) \rightarrow \operatorname{Mn}(s)+\mathrm{CO}_{2}(g)\) f. \(\mathrm{Sb}(s)+\mathrm{Cl}_{2}(g) \rightarrow \mathrm{SbCl}_{3}(s)\) g. \(\mathrm{CH}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightarrow \mathrm{CO}(g)+\mathrm{H}_{2}(g)\) h. \(\mathrm{FeS}(s)+\mathrm{HCl}(a q) \rightarrow \mathrm{FeCl}_{2}(a q)+\mathrm{H}_{2} \mathrm{~S}(g)\)

Short answer

The balanced chemical equations are: a. \(\mathrm{FeCl}_{3}(a q)+3\mathrm{KOH}(a q) \rightarrow \mathrm{Fe}(\mathrm{OH})_{3}(s)+3\mathrm{KCl}(a q)\) b. \(\mathrm{Pb}\left(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)_{2}(a q)+2\mathrm{KI}(a q) \rightarrow \mathrm{PbI}_{2}(s)+2\mathrm{KC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(a q)\) c. \(\mathrm{P}_{4} \mathrm{O}_{10}(s)+6\mathrm{H}_{2} \mathrm{O}(l) \rightarrow 4\mathrm{H}_{3} \mathrm{PO}_{4}(a q)\) d. \(\mathrm{Li}_{2} \mathrm{O}(s)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow 2\mathrm{LiOH}(a q)\) e. \(\operatorname{MnO}_{2}(s)+\mathrm{C}(s) \rightarrow \operatorname{Mn}(s)+\mathrm{CO}_{2}(g)\) f. \(2\mathrm{Sb}(s)+3\mathrm{Cl}_{2}(g) \rightarrow 2\mathrm{SbCl}_{3}(s)\) g. \(\mathrm{CH}_{4}(g)+2\mathrm{H}_{2} \mathrm{O}(g) \rightarrow \mathrm{CO}(g)+3\mathrm{H}_{2}(g)\) h. \(\mathrm{FeS}(s)+2\mathrm{HCl}(a q) \rightarrow \mathrm{FeCl}_{2}(a q)+\mathrm{H}_{2} \mathrm{~S}(g)\)

Step-by-step solution

Count the atoms on both sides of the equation

First, we need to count the number of atoms of each element on both the reactant and product side of the equation. Reactants: - Fe: 1 - Cl: 3 - K: 1 - O: 1 - H: 1 Products: - Fe: 1 - O: 3 - H: 3 - K: 1 - Cl: 1

Balance the atoms

We can see that the number of Cl and H atoms is not the same on both sides. To balance the equation, we need to adjust the coefficients in front of the chemical species. The balanced equation is: \(\mathrm{FeCl}_{3}(a q)+3\mathrm{KOH}(a q) \rightarrow \mathrm{Fe}(\mathrm{OH})_{3}(s)+3\mathrm{KCl}(a q)\) Now we have the same number of atoms for each element on both sides of the equation. #b. Balancing the equation \(\mathrm{Pb}\left(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)_{2}(a q)+\mathrm{KI}(a q) \rightarrow \mathrm{PbI}_{2}(s)+\mathrm{KC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(a q)\)#

Count the atoms on both sides of the equation

First, we need to count the number of atoms of each element on both the reactant and product side of the equation. Reactants: - Pb: 1 - C: 4 - H: 6 - O: 4 - K: 1 - I: 1 Products: - Pb: 1 - I: 2 - K: 1 - C: 2 - H: 3 - O: 2

Balance the atoms

We can see that the number of C, H, O, and I atoms is not the same on both sides. To balance the equation, we need to adjust the coefficients in front of the chemical species. The balanced equation is: \(\mathrm{Pb}\left(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)_{2}(a q)+2\mathrm{KI}(a q) \rightarrow \mathrm{PbI}_{2}(s)+2\mathrm{KC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(a q)\) Now we have the same number of atoms for each element on both sides of the equation. We follow a similar procedure for the remaining equations. #c. Balancing the equation \(\mathrm{P}_{4} \mathrm{O}_{10}(s)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{H}_{3} \mathrm{PO}_{4}(a q)\)# Balanced equation: \(\mathrm{P}_{4} \mathrm{O}_{10}(s)+6\mathrm{H}_{2} \mathrm{O}(l) \rightarrow 4\mathrm{H}_{3} \mathrm{PO}_{4}(a q)\) #d. Balancing the equation \(\mathrm{Li}_{2} \mathrm{O}(s)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{LiOH}(a q)\)# Balanced equation: \(\mathrm{Li}_{2} \mathrm{O}(s)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow 2\mathrm{LiOH}(a q)\) #e. Balancing the equation \(\operatorname{MnO}_{2}(s)+\mathrm{C}(s) \rightarrow \operatorname{Mn}(s)+\mathrm{CO}_{2}(g)\)# Balanced equation: \(\operatorname{MnO}_{2}(s)+\mathrm{C}(s) \rightarrow \operatorname{Mn}(s)+\mathrm{CO}_{2}(g)\) #f. Balancing the equation \(\mathrm{Sb}(s)+\mathrm{Cl}_{2}(g) \rightarrow \mathrm{SbCl}_{3}(s)\)# Balanced equation: \(2\mathrm{Sb}(s)+3\mathrm{Cl}_{2}(g) \rightarrow 2\mathrm{SbCl}_{3}(s)\) #g. Balancing the equation \(\mathrm{CH}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightarrow \mathrm{CO}(g)+\mathrm{H}_{2}(g)\)# Balanced equation: \(\mathrm{CH}_{4}(g)+2\mathrm{H}_{2} \mathrm{O}(g) \rightarrow \mathrm{CO}(g)+3\mathrm{H}_{2}(g)\) #h. Balancing the equation \(\mathrm{FeS}(s)+\mathrm{HCl}(a q) \rightarrow \mathrm{FeCl}_{2}(a q)+\mathrm{H}_{2} \mathrm{~S}(g)\)# Balanced equation: \(\mathrm{FeS}(s)+2\mathrm{HCl}(a q) \rightarrow \mathrm{FeCl}_{2}(a q)+\mathrm{H}_{2} \mathrm{~S}(g)\)

Key concepts

Stoichiometry

Stoichiometry is a foundational concept in chemistry that involves calculating the relative quantities of reactants and products in chemical reactions. It is based on the law of conservation of mass, which states that matter cannot be created or destroyed in a closed system. This principle allows us to predict how much of each substance is consumed or produced.
In every balanced chemical equation, stoichiometry helps to relate the moles of reactants to the moles of products. For instance, if an equation shows the reaction of 2 moles of hydrogen gas with 1 mole of oxygen gas to produce 2 moles of water, stoichiometry guides us to predict the proportions you need to react fully or the products you'll obtain.
Here are some ways stoichiometry is useful:

• Determining the exact amount of each reactant needed to produce a desired quantity of product.
• Predicting the amount of product formed in a reaction from a given amount of reactant.
• Calculating energy changes in reactions.

Understanding stoichiometry is critical because it allows chemists to design experiments with precision and predict outcomes accurately, avoiding waste of resources.

Chemical Reactions

Chemical reactions are processes where substances, known as reactants, are transformed into different substances, called products. During a reaction, bonds between atoms are broken and new ones are formed, leading to a change in the chemical properties and composition of the matter. Chemical reactions can be categorized into various types, such as synthesis, decomposition, single replacement, and double replacement reactions.
Each reaction is characterized by its specific reactants and products, as well as the conditions under which it occurs, like temperature, pressure, and catalysts involved. These factors can influence the rate of the reaction and the yield of products. In the classroom, understanding chemical reactions is vital because it forms the basis for further studies in chemistry, like kinetics and thermodynamics.
Key points of chemical reactions include:

• Conservation of mass: The mass of reactants equals the mass of products.
• Energy exchange: Reactions might absorb or release energy, usually in the form of heat (endothermic or exothermic reactions respectively).
• Reversibility: Some reactions can proceed in both forward and backward directions, leading to an equilibrium state.

Balancing Equations

Balancing chemical equations is crucial to correctly represent a chemical reaction. It ensures that the number of atoms for each element is the same on both sides of the equation, upholding the law of conservation of mass. Balancing equations involves adjusting coefficients, the numbers placed before compounds, not altering the molecular formula of the compounds themselves.
Basic steps involved in balancing chemical equations include:

• Write the unbalanced equation, identifying all reactants and products.
• Count the number of atoms for each element on both the reactant and product sides.
• Start by balancing elements that appear in only one reactant and one product.
• Balance remaining elements, adjusting coefficients as needed.
• Verify that all atoms balance and adjust coefficients if necessary.

Balancing equations helps to predict the proportions of substances involved in reactions, essential for laboratory settings to avoid wastage and ensure desired results. Understanding this concept solidifies a student's grasp of how reactions proceed in real-world scenarios.