Question

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. \(\mathrm{Ba}\left(\mathrm{ClO}_{3}\right)_{2}(s) \rightarrow \mathrm{BaCl}_{2}(s)+\mathrm{O}_{2}(s)\)

Short answer

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

Step-by-step solution

a. Balancing Fe3O4(s) + H2(g) → Fe(l) + H2O(g)

1. Balance oxygen atoms: Place a coefficient of 2 in front of H2O(g). \(\mathrm{Fe}_{3}\mathrm{O}_{4}(s)+\mathrm{H}_{2}(g) \rightarrow \mathrm{Fe}(l)+2\mathrm{H}_{2}\mathrm{O}(g)\) 2. Balance hydrogen atoms: Place a coefficient of 4 in front of H2(g). \(\mathrm{Fe}_{3}\mathrm{O}_{4}(s)+4\mathrm{H}_{2}(g) \rightarrow \mathrm{Fe}(l)+2\mathrm{H}_{2}\mathrm{O}(g)\) 3. Balance iron atoms: Place a coefficient of 3 in front of Fe(l). \(\mathrm{Fe}_{3}\mathrm{O}_{4}(s)+4\mathrm{H}_{2}(g) \rightarrow 3\mathrm{Fe}(l)+2\mathrm{H}_{2}\mathrm{O}(g)\)

b. Balancing K2SO4(aq) + BaCl2(aq) → BaSO4(s) + KCl(aq)

1. Balance barium and u... 2. Balance chlorine atoms: Place a coefficient of 2 in front of KCl(aq). \(\mathrm{K}_{2}\mathrm{SO}_{4}(a q)+\mathrm{BaCl}_{2}(a q) \rightarrow \mathrm{BaSO}_{4}(s)+2\mathrm{KCl}(a q)\)

c. Balancing HCl(aq) + FeS(s) → FeCl2(aq) + H2S(g)

1. Balance hydrogen atoms: Place a coefficient of 2 in front of HCl(aq). 2. Balance chlorine atoms: Place a coefficient of 2 in front of HCl(aq).

d. Balancing Br2(g) + H2O(l) + SO2(g) → HBr(aq) + H2SO4(aq)

1. Balance hydrogen atoms: Place a coefficient of 2 in front of HBr(aq). 2. Balance selenium atoms: Place a coefficient of 2 in front of Br2(g). 3. Balance bromine atoms: Place a coefficient of 2 in front of HBr(aq).

e. Balancing CS2(l) + Cl2(g) → CCl4(l) + S2Cl2(g)

1. Balance carbon atoms: Place a coefficient of 4 in front of Cl2(g). 2. Balance sulfur atoms: Place a coefficient of 2 in front of S2Cl2(g).

f. Balancing Cl2O7(g) + Ca(OH)2(aq) → Ca(ClO4)2(aq) + H2O(l)

1. Balance oxygen atoms: Place a coefficient of 2 in front of H2O(l). 2. Balance hydrogen atoms: Place a coefficient of 2 in front of H2O(l).

g. Balancing PBr3(l) + H2O(l) → H3PO3(aq) + HBr(g)

1. Balance hydrogen atoms: Place a coefficient of 3 in front of HBr(g). 2. Balance bromine atoms: Place a coefficient of 3 in front of HBr(g).

h. Balancing Ba(ClO3)2(s) → BaCl2(s) + O2(s)

1. Balance oxygen atoms: Place a coefficient of 2 in front of O2(s).

Key concepts

Stoichiometry

Stoichiometry can be likened to a recipe for a chemical reaction, providing the exact proportion of reactants needed to form desired products. It is rooted in the conservation of mass, where the total mass of reactants equals the total mass of products.

Delving deeper, stoichiometry deals with the quantitative relationships between the substances involved in a chemical reaction. It includes calculating the amounts of reactants or products that are consumed or produced using the balanced chemical equations. An essential part of solving stoichiometry problems is being able to balance a chemical equation which reflects these mass relationships between reactants and products.

Chemical Reactions

A chemical reaction is a process where reactants are transformed into products through the breaking and forming of chemical bonds. To describe these reactions, chemical equations are used, which consist of reactant(s) on the left side and product(s) on the right.

Each reaction embodies specific characteristics, such as synthesis, decomposition, single replacement, or double replacement. For example, in the textbook exercise on balancing equations, different types of reactions are showcased. To balance a chemical equation correctly, one must understand these types of reactions as they provide hints on how different elements are likely to combine or decompose, which is critical for determining the correct coefficients.

Coefficients in Chemical Equations

Coefficients in chemical equations play a pivotal role in ensuring the law of conservation of mass is satisfied. They indicate the number of units of each substance that participate in the reaction. Adjusting these numerical coefficients is how we balance chemical equations; it is not permissible to alter the chemical formulas of the reactants or products.

In the step-by-step solutions provided in the textbook exercise, coefficients are strategically placed to ensure that the number of atoms of each element is the same on both sides of the equation. Hence, in the first example with iron(III) oxide and hydrogen reacting to form iron and water, the coefficients ensure there are equal numbers of iron, oxygen, and hydrogen atoms on both sides, following the balanced equation \( \mathrm{Fe}_3\mathrm{O}_4(s)+4\mathrm{H}_2(g) \rightarrow 3\mathrm{Fe}(l)+2\mathrm{H}_2\mathrm{O}(g) \).

The mastery of balancing chemical equations is paramount in stoichiometry as it underpins the ability to predict the outcomes of reactions and to perform quantitative calculations required for many practical applications in chemistry.