Question

Balance each of the following equations that describe precipitation reactions. a. \(\mathrm{Na}_{2} \mathrm{SO}_{4}(a q)+\mathrm{CaCl}_{2}(a q) \rightarrow \mathrm{CaSO}_{4}(s)+\mathrm{NaCl}(a q)\) b. \(\mathrm{Co}\left(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)_{2}(a q)+\mathrm{Na}_{2} \mathrm{~S}(a q) \rightarrow \mathrm{CoS}(s)+\mathrm{NaC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(a q)\) c. \(\mathrm{KOH}(a q)+\mathrm{NiCl}_{2}(a q) \rightarrow \mathrm{Ni}(\mathrm{OH})_{2}(s)+\mathrm{KCl}(a q)\)

Short answer

The balanced equations for the given precipitation reactions are: a. \(\mathrm{Na}_{2} \mathrm{SO}_{4}(a q)+\mathrm{CaCl}_{2}(a q) \rightarrow \mathrm{CaSO}_{4}(s)+2 \mathrm{NaCl}(a q)\) b. \(\mathrm{Co}\left(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)_{2}(a q)+\mathrm{Na}_{2} \mathrm{~S}(a q) \rightarrow \mathrm{CoS}(s)+2 \mathrm{NaC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(a q)\) c. \(2\mathrm{KOH}(a q)+\mathrm{NiCl}_{2}(a q) \rightarrow \mathrm{Ni}(\mathrm{OH})_{2}(s)+2 \mathrm{KCl}(a q)\)

Step-by-step solution

Write down the unbalanced equation

The unbalanced equation for the first reaction is given as: \(\mathrm{Na}_{2} \mathrm{SO}_{4}(a q)+\mathrm{CaCl}_{2}(a q) \rightarrow \mathrm{CaSO}_{4}(s)+\mathrm{NaCl}(a q)\)

Balance the equation

To balance this equation, we will first write down the number of atoms of each element on both sides of the equation: - Na: 2 on the left, 1 on the right - S: 1 on the left, 1 on the right - O: 4 on the left, 4 on the right - Ca: 1 on the left, 1 on the right - Cl: 2 on the left, 1 on the right Notice that we have an imbalance in the number of Na and Cl atoms. To balance the Na atoms, we can put a coefficient of 2 in front of NaCl: now we have 2 Na atoms on both sides. To balance the Cl atoms, no changes are needed since NaCl already has a coefficient of 2, so there are 2 Cl atoms on both sides now. The balanced equation is: \(\mathrm{Na}_{2} \mathrm{SO}_{4}(a q)+\mathrm{CaCl}_{2}(a q) \rightarrow \mathrm{CaSO}_{4}(s)+2 \mathrm{NaCl}(a q)\) #b. Balancing the second equation#

Write down the unbalanced equation

The unbalanced equation for the second reaction is given as: \(\mathrm{Co}\left(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)_{2}(a q)+\mathrm{Na}_{2} \mathrm{~S}(a q) \rightarrow \mathrm{CoS}(s)+\mathrm{NaC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(a q)\)

Balance the equation

To balance this equation, we will do the same as in the first equation: - Co: 1 on the left, 1 on the right - C: 4 on the left, 2 on the right - H: 6 on the left, 3 on the right - O: 4 on the left, 2 on the right - Na: 2 on the left, 1 on the right - S: 1 on the left, 1 on the right Notice that we have an imbalance in the number of C, H, O, and Na atoms. To balance the C, H, and O atoms, we can put a coefficient of 2 in front of NaC2H3O2. This gives us 4 C atoms, 6 H atoms, and 4 O atoms on both sides of the equation. To balance the Na atoms, no changes are needed, since NaC2H3O2 already has a coefficient of 2. The balanced equation is: \(\mathrm{Co}\left(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)_{2}(a q)+\mathrm{Na}_{2} \mathrm{~S}(a q) \rightarrow \mathrm{CoS}(s)+2 \mathrm{NaC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(a q)\) #c. Balancing the third equation#

Write down the unbalanced equation

The unbalanced equation for the third reaction is given as: \(\mathrm{KOH}(a q)+\mathrm{NiCl}_{2}(a q) \rightarrow \mathrm{Ni}(\mathrm{OH})_{2}(s)+\mathrm{KCl}(a q)\)

Balance the equation

To balance this equation, we will do the same as before: - K: 1 on the left, 1 on the right - O: 1 on the left, 2 on the right - H: 1 on the left, 2 on the right - Ni: 1 on the left, 1 on the right - Cl: 2 on the left, 1 on the right Notice that we have an imbalance in the number of O, H, and Cl atoms. To balance the O and H atoms, we can put a coefficient of 2 in front of KOH. This gives us 2 O atoms and 2 H atoms on both sides of the equation. To balance the Cl atoms, we can put a coefficient of 2 in front of KCl. This gives us 2 Cl atoms on both sides of the equation. The balanced equation is: \(2\mathrm{KOH}(a q)+\mathrm{NiCl}_{2}(a q) \rightarrow \mathrm{Ni}(\mathrm{OH})_{2}(s)+2 \mathrm{KCl}(a q)\)

Key concepts

Precipitation Reactions

Precipitation reactions occur when two aqueous solutions combine, resulting in the formation of an insoluble solid, known as a precipitate. These reactions are a vital part of chemistry because they allow us to separate components in a solution.
When two solutions containing ionic compounds are mixed, the ions may rearrange themselves, and if an insoluble compound is formed, it appears as a solid precipitate separating from the solution.

• For example, when mixing sodium sulfate (\( \mathrm{Na}_2 \mathrm{SO}_4 \)) and calcium chloride (\( \mathrm{CaCl}_2 \)), calcium sulfate (\( \mathrm{CaSO}_4 \)) forms as a precipitate.

Recognizing precipitation reactions involves understanding the solubility rules, which predict the solubility of ionic compounds in water. By knowing these rules, it is possible to predict whether a precipitate will form in a given chemical reaction.

Stoichiometry

Stoichiometry is the branch of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction.
Understanding stoichiometry allows us to predict how much product forms from given reactants or how much reactant is needed to form a desired quantity of product.
To perform stoichiometric calculations in a balanced precipitation reaction, consider these steps:

• First, write the balanced chemical equation. Balancing ensures the same number of each type of atom on both sides of the equation.
• Determine the mol ratio from the balanced equation, which allows for conversion between moles of reactants and products. This ratio is crucial for calculating theoretical yields and reacting amounts.

By having these balanced equations, you can confidently carry out all calculations involving mass, moles, and volume in chemical reactions.

Chemical Reactions

Chemical reactions involve the breaking and forming of bonds to convert reactants into products. They are fundamental to understanding how substances interact and transform.
For meaningful analysis and problem-solving, every chemical reaction should be balanced. In the process of balancing, we adjust coefficients so that the number of each type of atom is the same on either side of the equation.
There are different types of reactions, among which precipitation is one such type, crucial in producing various substances. Here are some key points:

• Reactants: Substances that undergo change (e.g., in our exercise, sodium sulfate and calcium chloride).
• Products: New substances formed (e.g., calcium sulfate precipitate).
• Balancing Equations: Ensures mass conservation and correct reactive proportions.

Each reaction type serves a unique purpose in chemistry, whether forming a brand new compound or isolating a desired product.

Aqueous Solutions

Aqueous solutions are solutions where water is the solvent. In chemistry, many reactions occur in aqueous solutions, making them ubiquitous and important.
When a substance dissolves in water, it disassembles into ions allowing it to become part of an aqueous solution.

• For example, sodium sulfate (\( \mathrm{Na}_2 \mathrm{SO}_4 \)) dissolves into \( 2 \mathrm{Na}^+ \) and \( \mathrm{SO}_4^{2-} \).

Aqueous solutions enable ionic reactions, important for causing precipitation and other types of reactions.
By being part of an aqueous solution, reactants can move freely, colliding and reacting at much higher speeds compared to solid-phase reactions.
Understanding the behavior of ions in aqueous solutions is crucial to predicting which reactions will occur and what products will form, thus deepening our understanding of chemistry on the molecular level.