Question

Balance each of the following chemical equations. a. \(\mathrm{Cl}_{2}(g)+\mathrm{KBr}(a q) \rightarrow \mathrm{Br}_{2}(l)+\mathrm{KCl}(a q)\) b. \(\operatorname{Cr}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{Cr}_{2} \mathrm{O}_{3}(s)\) c. \(\mathrm{P}_{4}(s)+\mathrm{H}_{2}(g) \rightarrow \mathrm{PH}_{3}(g)\) d. \(\mathrm{Al}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q)+\mathrm{H}_{2}(g)\) e. \(\mathrm{PCl}_{3}(l)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{H}_{3} \mathrm{PO}_{3}(a q)+\mathrm{HCl}(a q)\) \(\mathrm{f} \cdot \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightarrow \mathrm{SO}_{3}(g)\) g. \(\mathrm{C}_{7} \mathrm{H}_{16}(l)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) h. \(\mathrm{C}_{2} \mathrm{H}_{6}(g)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\)

Short answer

a. \( \mathrm{Cl}_{2}(g) + 2\mathrm{KBr}(aq) \rightarrow \mathrm{Br}_{2}(l) + 2\mathrm{KCl}(aq) \) b. \( 2\mathrm{Cr}(s) + 3\mathrm{O}_{2}(g) \rightarrow 2\mathrm{Cr}_{2}\mathrm{O}_{3}(s) \) c. \( \mathrm{P}_{4}(s) + 6\mathrm{H}_{2}(g) \rightarrow 4\mathrm{PH}_{3}(g) \) d. \( 2\mathrm{Al}(s) + 3\mathrm{H}_{2}\mathrm{SO}_{4}(aq) \rightarrow \mathrm{Al}_{2}(\mathrm{SO}_{4})_{3}(aq) + 6\mathrm{H}_{2}(g) \) e. \( \mathrm{PCl}_{3}(l) + 3\mathrm{H}_{2}\mathrm{O}(l) \rightarrow \mathrm{H}_{3}\mathrm{PO}_{3}(aq) + 3\mathrm{HCl}(aq)\) f. \( 2\mathrm{SO}_{2}(g) + \mathrm{O}_{2}(g) \rightarrow 2\mathrm{SO}_{3}(g) \) g. \( \mathrm{C}_{7}\mathrm{H}_{16}(l) + 11\mathrm{O}_{2}(g) \rightarrow 7\mathrm{CO}_{2}(g) + 8\mathrm{H}_{2}\mathrm{O}(g) \) h. \( 2\mathrm{C}_{2}\mathrm{H}_{6}(g) + 7\mathrm{O}_{2}(g) \rightarrow 4\mathrm{CO}_{2}(g) + 6\mathrm{H}_{2}\mathrm{O}(g) \)

Step-by-step solution

Identify the elements

In this equation, we have four elements: Cl, K, Br, and another Cl.

Balance the atoms

We notice that the second Cl is combined with K. So, we'll look at Cl, K, and Br atoms first. - Cl: There are 2 Cl atoms on the left and 1 Cl atom on the right. To balance the Cl atoms, we need to have 2 Cl atoms on the right as well, so we add a coefficient of 2 in front of KCl: \(\mathrm{Cl}_{2}(g)+\mathrm{KBr}(a q) \rightarrow \mathrm{Br}_{2}(l)+2\mathrm{KCl}(a q)\) - K: There is 1 K atom on the left and 2 K atoms on the right. Since the equation is already balanced, no further adjustments are required. - Br: There are 1 Br atoms on the left and 2 Br atoms on the right. To balance the Br atoms, we can add a coefficient of 2 in front of KBr: \(\mathrm{Cl}_{2}(g)+2\mathrm{KBr}(a q) \rightarrow \mathrm{Br}_{2}(l)+2\mathrm{KCl}(a q)\) Now the equation is balanced. #b. Balancing the equation \(\operatorname{Cr}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{Cr}_{2} \mathrm{O}_{3}(s)\)#:

Identify the elements

In this equation, we have two elements: Cr and O.

Balance the atoms

We'll balance Cr and O atoms: - Cr: There is 1 Cr atom on the left and 2 Cr atoms on the right. To balance the Cr atoms, we can add a coefficient of 2 in front of Cr: \(2\operatorname{Cr}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{Cr}_{2} \mathrm{O}_{3}(s)\) - O: There are 2 O atoms on the left and 3 O atoms on the right. To balance the O atoms, we need to have 6 O atoms on both sides, so we add a coefficient of 3 in front of O2 on the left and a coefficient of 2 in front of Cr2O3 on the right: \(2\operatorname{Cr}(s)+3\mathrm{O}_{2}(g) \rightarrow 2\mathrm{Cr}_{2} \mathrm{O}_{3}(s)\) Now the equation is balanced. I will now balance the remaining equations using the same approach as above. #c. Balancing the equation \(\mathrm{P}_{4}(s)+\mathrm{H}_{2}(g) \rightarrow \mathrm{PH}_{3}(g)\)#: Step 1: Identify the elements: P and H. Step 2: Balance the atoms: - P: 4 P atoms on the left. Add a coefficient of 4 in front of PH3: \(\mathrm{P}_{4}(s)+\mathrm{H}_{2}(g) \rightarrow 4\mathrm{PH}_{3}(g)\) - H: 2 H atoms on the left, 12 H atoms on the right. Add a coefficient of 6 in front of H2: \(\mathrm{P}_{4}(s)+6\mathrm{H}_{2}(g) \rightarrow 4\mathrm{PH}_{3}(g)\) Now the equation is balanced. #d. Balancing the equation \(\mathrm{Al}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q)+\mathrm{H}_{2}(g)\)#: Step 1: Identify the elements: Al, H, S, and O. Step 2: Balance the atoms: - Al: 1 Al atom on the left, 2 Al atoms on the right. Add a coefficient of 2 in front of Al: \(2\mathrm{Al}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q)+\mathrm{H}_{2}(g)\) - H: 2 H atoms on the left and the right. No changes needed. - S: 1 S atom on the left, 3 S atoms on the right. Add a coefficient of 3 in front of H2SO4: \(2\mathrm{Al}(s)+3\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q)+\mathrm{H}_{2}(g)\) - O: 12 O atoms on the left, 12 O atoms on the right. No changes needed. - Balancing the H2 on the right side: Add a coefficient of 6 in front of H2: \(2\mathrm{Al}(s)+3\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q)+6\mathrm{H}_{2}(g)\) Now the equation is balanced. And so on for the remaining equations.

Key concepts

Chemical Reactions

Understanding chemical reactions is key to grasping the fundamental principles of chemistry. A chemical reaction is a process that leads to the transformation of one set of chemical substances to another. During these reactions, the bonds between atoms break or form, leading to the creation of new substances with different properties from those of the reactants.

When we write out a chemical reaction, we use a chemical equation, which illustrates the substances involved in the reaction. Reactants, found on the left side of the equation, interact to produce the products, listed on the right. The arrow in the equation symbolizes the direction of the reaction. An important law that governs these reactions is the Law of Conservation of Mass, which states that matter is neither created nor destroyed. Thus, the mass of the reactants must equal the mass of the products. This law is the foundation of balancing chemical equations—as you noticed when we added coefficients in front of each substance to ensure the same number of each type of atom on both sides of the equation.

Stoichiometry

Stoichiometry is a branch of chemistry that deals with the quantitative aspects of chemical reactions. It helps chemists understand the relationships between the reactants and products in a reaction, allowing for predictions about the amount of substances produced or required.

Crucial to this concept is the mole, a unit that measures the amount of substance. A mole contains Avogadro's number of particles (\(6.022 \times 10^{23}\)), which is similar to how a dozen refers to twelve items. Stoichiometry relies on balanced chemical equations because they provide the ratios of the moles of reactants and products. These ratios are called stoichiometric coefficients, and they are essential in problem-solving when it comes to predicting yield or determining required reactant quantities. For instance, in the reaction \(2\mathrm{H}_2 + \mathrm{O}_2 \rightarrow 2\mathrm{H}_2\mathrm{O}\), the stoichiometry of the reaction indicates that two moles of hydrogen gas react with one mole of oxygen gas to produce two moles of water.

Chemical Formulas

Chemical formulas represent the composition of a molecule, indicating the elements present and the number of atoms of each element. For example, in the chemical formula \(\mathrm{H}_2\mathrm{O}\), the elements hydrogen (H) and oxygen (O) are present, with two hydrogen atoms and one oxygen atom making up a water molecule.

There are several types of chemical formulas, including the empirical formula, which shows the simplest whole-number ratio of atoms in a compound, and the molecular formula, which gives the actual number of atoms in a molecule. For our balancing equations exercise, we deal mostly with molecular formulas. Understanding these formulas is important for balancing chemical equations because they tell us the number of each atom that must be accounted for on both sides of the equation. Correctly interpreting chemical formulas is essential to achieve the proper balance and therefore, the correct stoichiometry in chemical reactions.