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. \(\operatorname{Sb}(s)+\mathrm{Cl}_{2}(g) \rightarrow \mathrm{SbCl}_{3}(s)\) g. \(\mathrm{CH}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightarrow \mathrm{CO}(g)+\mathrm{H}_{2}(g)\) h. \(\operatorname{FeS}(s)+\operatorname{HCl}(a q) \rightarrow \operatorname{Fe} \mathrm{Cl}_{2}(a q)+\mathrm{H}_{2} \mathrm{S}(g)\)

Short answer

The balanced chemical equations are as follows: 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. \( \mathrm{CH}_{4}(g)+2\mathrm{H}_{2} \mathrm{O}(g) \rightarrow\mathrm{CO}(g)+4\mathrm{H}_{2}(g) \) h. \( \operatorname{FeS}(s)+2\operatorname{HCl}(a q) \rightarrow \operatorname{Fe} \mathrm{Cl}_{2}(a q)+\mathrm{H}_{2} \mathrm{S}(g) \)

Step-by-step solution

a. Balancing FeCl3(aq) + KOH(aq) -> Fe(OH)3(s) + KCl(aq)

To balance this equation, we need to make sure that the number of atoms of each element is equal on both sides of the equation. Let's count the number of atoms for each element: - Fe: 1 on both sides - Cl: 3 on the left side, 1 on the right side - K: 1 on both sides - O: 1 on the left side, 3 on the right side - H: 1 on the left side, 3 on the right side To balance the equation, we multiply KCl(aq) by 3 on the right side, and multiply KOH(aq) by 3 on the left side. The balanced equation is: \( \mathrm{FeCl}_{3}(a q)+3 \mathrm{KOH}(a q) \rightarrow \mathrm{Fe}(\mathrm{OH})_{3}(s)+3 \mathrm{KCl}(a q) \)

b. Balancing Pb(C2H3O2)2(aq) + KI(aq) -> PbI2(s) + KC2H3O2(aq)

Let's count the number of atoms for each element: - Pb: 1 on both sides - C: 4 on the left side, 2 on the right side - H: 6 on the left side, 3 on the right side - O: 4 on the left side, 2 on the right side - K: 1 on the right side - I: 1 on the left side, 2 on the right side To balance this equation, we multiply KC2H3O2(aq) by 2 on the right side. The balanced equation is: \( \mathrm{Pb}\left(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)_{2}(a q)+2\mathrm{KI}(a q) \rightarrow \mathrm{PbI}_{2}(s)+2\mathrm{KC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(a q) \)

c. Balancing P4O10(s) + H2O(l) -> H3PO4(aq)

Let's count the number of atoms for each element: - P: 4 on the left side, 1 on the right side - O: 10 on the left side, 4 on the right side - H: 2 on the left side, 3 on the right side To balance this equation, we multiply H3PO4(aq) by 4 on the right side and H2O(l) by 6 on the left side. The balanced equation is: \( \mathrm{P}_{4} \mathrm{O}_{10}(s)+6\mathrm{H}_{2} \mathrm{O}(l) \rightarrow 4\mathrm{H}_{3} \mathrm{PO}_{4}(a q) \)

d. Balancing Li2O(s) + H2O(l) -> LiOH(aq)

Let's count the number of atoms for each element: - Li: 2 on the left side, 1 on the right side - O: 1 on both sides - H: 2 on the left side, 1 on the right side To balance this equation, we multiply LiOH(aq) by 2 on the right side. The balanced equation is: \( \mathrm{Li}_{2} \mathrm{O}(s)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow 2\mathrm{LiOH}(a q) \)

e. Balancing MnO2(s) + C(s) -> Mn(s) + CO2(g)

Let's count the number of atoms for each element: - Mn: 1 on both sides - O: 2 on the left side, 2 on the right side - C: 1 on both sides The equation is already balanced: \( \operatorname{MnO}_{2}(s)+\mathrm{C}(s) \rightarrow \operatorname{Mn}(s)+\mathrm{CO}_{2}(g) \)

f. Balancing Sb(s) + Cl2(g) -> SbCl3(s)

Let's count the number of atoms for each element: - Sb: 1 on both sides - Cl: 2 on the left side, 3 on the right side To balance this equation, multiply Cl2(g) by 3/2 on the left side and SbCl3(s) by 2 on the right side: \( 2\mathrm{Sb}(s)+3\mathrm{Cl}_{2}(g) \rightarrow 2\mathrm{SbCl}_{3}(s) \)

g. Balancing CH4(g) + H2O(g) -> CO(g) + H2(g)

Let's count the number of atoms for each element: - C: 1 on both sides - H: 4 on the left side, 2 on the right side - O: 1 on both sides To balance this equation, multiply H2(g) by 4 on the right side and H2O(g) by 2 on the left side: \( \mathrm{CH}_{4}(g)+2\mathrm{H}_{2} \mathrm{O}(g) \rightarrow\mathrm{CO}(g)+4\mathrm{H}_{2}(g) \)

h. Balancing FeS(s) + HCl(aq) -> FeCl2(aq) + H2S(g)

Let's count the number of atoms for each element: - Fe: 1 on both sides - S: 1 on both sides - H: 1 on the left side, 2 on the right side - Cl: 1 on the left side, 2 on the right side To balance this equation, we multiply HCl(aq) by 2 on the left side The balanced equation is: \( \operatorname{FeS}(s)+2\operatorname{HCl}(a q) \rightarrow \operatorname{Fe} \mathrm{Cl}_{2}(a q)+\mathrm{H}_{2} \mathrm{S}(g) \)

Key concepts

Stoichiometry

Stoichiometry is akin to a recipe for a chemical reaction. It allows chemists to calculate the precise amounts of reactants required to produce desired products. At its core, stoichiometry is based on the law of conservation of mass, which states that in a chemical reaction, matter is neither created nor destroyed. This means all atoms present in the reactants must be accounted for in the products.

In practical terms, stoichiometry provides the mole ratios needed to balance chemical equations, representing how many particles of one substance react with another. Since reactions often occur between substances in different physical states, such as solids, liquids, and gases, stoichiometry is vital in determining how these different materials will interact in terms of moles.

An example from the exercise involves balancing the reaction between iron(III) chloride and potassium hydroxide to form iron(III) hydroxide and potassium chloride. Through stoichiometry, we determined that for every mole of iron(III) chloride, we need three moles of potassium hydroxide to yield three moles of potassium chloride and one mole of iron(III) hydroxide, ensuring mass is conserved.

It's crucial for students to grasp the method of balancing equations by using stoichiometric coefficients - the numbers placed before compounds in a chemical equation to indicate the amount of moles. These coefficients are the key to understanding the quantitative aspect of chemical reactions.

Chemical Reactions

Chemical reactions are processes where substances, the reactants, transform into different substances, the products. When we express this transformation, we write a chemical equation that should be balanced to reflect the same number of atoms of each element on both sides. This is due to the principle mentioned earlier - matter is conserved during a reaction.

There are different types of chemical reactions such as synthesis, decomposition, single replacement, and double replacement—all involving the breaking and forming of chemical bonds. For instance, the reaction between methane and water in the exercise is a type of double replacement reaction.

Understanding chemical equations is crucial, not just for theoretical discussions but also for practical applications like predicting the products of a reaction or determining the amounts needed for a reaction to proceed completely. In classroom settings, improving proficiency with chemical reactions involves practicing with various equations and recognizing patterns of how different substance classes interact, such as acids with bases, metals with nonmetals, and hydrocarbons with oxygen.

Mole Concept

The mole concept is a bridge between the microscopic world of atoms and molecules and the macroscopic world we interact with daily. One mole is defined as the amount of substance that contains as many entities (atoms, molecules, ions, or other particles) as there are atoms in 12 grams of carbon-12. This number is known as Avogadro's number, approximately 6.022 x 10^23 entities in one mole.

In the exercise, when we balance equations, we are essentially dealing with moles. Even though we cannot count out atoms individually, using the mole allows us to measure out amounts of substances that will have a specific number of particles. This is why the mole concept is pivotal in stoichiometry — it is the foundation that relates the mass of a substance to the number of particles it contains.

Mastering the mole concept entails understanding the relationships between mass, moles, and the number of particles. It involves being comfortable with conversions between these units, which is critical for tasks like determining how many molecules are in a given sample, or how much of a reactant is needed to fully complete a reaction.