Question

Complete and balance each equation. If no reaction occurs, write NO REACTION. (a) \(\mathrm{NaOH}(a q)+\mathrm{FeBr}_{3}(a q)\) (b) \(\mathrm{BaCl}_{2}(a q)+\mathrm{AgNO}_{3}(a q)\) (c) \(\mathrm{Na}_{2} \mathrm{CO}_{3}(a q)+\mathrm{CoCl}_{2}(a q) \longrightarrow\) (d) \(\mathrm{K}_{2} \mathrm{~S}(a q)+\mathrm{BaCl}_{2}(a q) \longrightarrow\)

Short answer

The balanced equations are (a) 3 NaOH + FeBr3 → Fe(OH)3 + 3 NaBr, (b) BaCl2 + 2 AgNO3 → 2 AgCl + Ba(NO3)2, (c) Na2CO3 + CoCl2 → CoCO3 + 2 NaCl, (d) NO REACTION.

Step-by-step solution

Predict the Products for Reaction (a)

The reaction between NaOH (a strong base) and FeBr3 (an iron(III) bromide solution) will undergo double displacement to produce Fe(OH)3 and NaBr. Iron(III) hydroxide forms a precipitate, while sodium bromide remains in solution.

Balance Reaction (a)

Balance the atoms and charges. The balanced chemical equation for reaction (a) is: \[3 \mathrm{NaOH}(aq) + \mathrm{FeBr}_{3}(aq) \longrightarrow \mathrm{Fe(OH)}_{3}(s) + 3 \mathrm{NaBr}(aq)\]

Predict the Products for Reaction (b)

Barium chloride reacts with silver nitrate to undergo double displacement, with the insoluble silver chloride and soluble barium nitrate as the products.

Balance Reaction (b)

Ensure each element has equal atoms on both sides of the equation. The balance equation for reaction (b) is: \[\mathrm{BaCl}_{2}(aq) + 2 \mathrm{AgNO}_{3}(aq) \longrightarrow 2 \mathrm{AgCl}(s) + \mathrm{Ba(NO}_{3})_{2}(aq)\]

Predict the Products for Reaction (c)

Sodium carbonate reacts with cobalt(II) chloride resulting in the double displacement to form sodium chloride (NaCl) and cobalt(II) carbonate (CoCO3), with the latter often being a precipitate.

Balance Reaction (c)

Adjust the coefficients to balance the equation. The balanced equation for reaction (c) is: \[\mathrm{Na}_{2} \mathrm{CO}_{3}(aq) + \mathrm{CoCl}_{2}(aq) \longrightarrow \mathrm{CoCO}_{3}(s) + 2 \mathrm{NaCl}(aq)\]

Predict the Products for Reaction (d)

Potassium sulfide reacts with barium chloride yielding double displacement to form potassium chloride (KCl) and barium sulfide (BaS). However, BaS is soluble in water and will not form a precipitate, leading to the same ions in solution and therefore no reaction.

State That Reaction (d) Yields No Reaction

No reaction occurs because the potential products are soluble. Write NO REACTION for reaction (d).

Key concepts

Double Displacement Reaction

Understanding double displacement reactions is essential in the field of chemistry. These reactions occur when the constituent ions in two compounds switch places with each other, resulting in the formation of two new compounds. It's like a dance where two couples swap partners. A typical characteristic of this reaction type is the exchange of cations and anions between two reacting chemical species.

For example, in the solution for reaction (a), \(\mathrm{NaOH}(aq) + \mathrm{FeBr}_3(aq)\), sodium (Na+) from NaOH and iron (Fe3+) from FeBr3 swap their respective anion partners. This results in the formation of Fe(OH)3 and NaBr. The (aq) notation indicates that these compounds are in an aqueous solution, meaning they are dissolved in water. These reactions often occur in solutions and are a staple in laboratory and industrial processes.

This interchanging of ions follows a predictable pattern, often simplifying the prediction of the products of a reaction. Knowing the reactivity and the soluble or insoluble nature of the products helps in determining the outcome of a double displacement reaction.

Solubility and Precipitation

Solubility and precipitation are interconnected concepts within the context of double displacement reactions. Solubility refers to the ability of a substance to dissolve in a solvent, such as water. In contrast, a precipitate is an insoluble solid that emerges from a solution when two solutes react.

In the textbook solutions provided, we encounter instances where these concepts play a crucial role. For instance, Fe(OH)3 and AgCl are precipitates in reactions (a) and (b) respectively, indicating that they do not dissolve in water but rather form a solid. Conversely, compounds like NaBr and Ba(NO3)2 remain dissolved and do not precipitate.

The solubility rules help us predict whether a substance will be soluble or not. Compounds containing nitrates and most salts with sodium and potassium ions are typically soluble. Conversely, carbonates, phosphates, and hydroxides often precipitate, unless they are partnered with highly soluble cations.

Therefore, understanding the solubility of compounds is key to predicting the formation of a precipitate in a chemical reaction. It is also useful for figuring out which compounds will remain in the solution after the reaction has taken place.

Writing Chemical Equations

Writing chemical equations is a foundational skill in chemistry, as it represents the shorthand communication of a chemical reaction. It involves not only identifying the reactants and products but also ensuring the law of conservation of mass is obeyed — the atoms of each element should be balanced on both sides of the equation.

For clear communication, certain conventions are observed, such as writing the state of each compound (solid \(s\), liquid \(l\), gas \(g\), or aqueous \(aq\)) next to it. The step-by-step solutions above meticulously demonstrate this with balanced equations for reactions (a), (b), and (c), and the declaration of no reaction for (d).

When writing these equations, one starts by predicting the products based on the reactants’ chemical nature and solubility rules. Then, coefficients are adjusted to balance the equation, ensuring the same number of each type of atom on both sides. For instance, the balanced equation from reaction (a), \[3 \mathrm{NaOH}(aq) + \mathrm{FeBr}_3(aq) \longrightarrow \mathrm{Fe(OH)}_3(s) + 3 \mathrm{NaBr}(aq)\], shows the exact stoichiometric balance needed to reflect the quantities of reactants and products.

Learning to write balanced chemical equations is crucial for anyone studying chemistry, as it provides a clear picture of the chemical processes involved and is the first step towards quantitative chemical analysis.