Question

Complete and balance each equation. If no reaction occurs, write NO REACtION. a. \(\operatorname{Lil}(a q)+\mathrm{BaS}(a q) \longrightarrow\) b. \(\mathrm{KCl}(a q)+\mathrm{CaS}(a q) \longrightarrow\) c. \(\operatorname{CrBr}_{2}(a q)+\mathrm{Na}_{2} \mathrm{CO}_{3}(a q) \longrightarrow\) d. \(\mathrm{NaOH}(a q)+\mathrm{FeCl}_{3}(a q) \longrightarrow\)

Short answer

a. NO REACTION b. NO REACTION c. \(\mathrm{CrBr}_{2}(aq) + \mathrm{Na}_{2}CO_{3}(aq) \longrightarrow \mathrm{CrCO}_{3}(s) + 2\mathrm{NaBr}(aq)\) d. \(3\mathrm{NaOH}(aq) + \mathrm{FeCl}_{3}(aq) \longrightarrow \mathrm{Fe(OH)}_{3}(s) + 3\mathrm{NaCl}(aq)\)

Step-by-step solution

- Understanding the Reactivity

First, it is important to understand if a reaction takes place by considering the solubility rules and possible products that could form. The products can form if they are either a precipitate (solid), a gas that escapes, or a weakly ionizing molecule like water.

- Predicting the Products

For each reaction, use the concept of double displacement, where the anions and cations from the reactants switch partners. Determine whether the new combinations are soluble or form a precipitate.

- Writing the Products for Reaction a

For the reaction of LiI and BaS, exchange the ions: Li goes with S and Ba goes with I, giving possible products of Li2S and BaI2. According to solubility rules, both of these compounds are soluble, so no reaction occurs.

- Writing the Products for Reaction b

For KCl and CaS, switch the ions: K goes with S and Ca with Cl, yielding possible products K2S and CaCl2. Since both are soluble in water, no reaction occurs.

- Writing the Products for Reaction c

For CrBr2 and Na2CO3, we exchange ions: Cr pairs with CO3, and Na pairs with Br, producing Cr(CO3)2 and NaBr. Sodium bromide, NaBr, is soluble, but chromium(II) carbonate, Cr(CO3)2, is insoluble, indicating a reaction takes place, forming a precipitate.

- Balancing the Equation for Reaction c

Write the balanced equation considering the insoluble product: Cr(CO3)2. Because there are 2 sodium ions for every carbonate ion, and 2 bromide ions for every chromium ion, the balanced equation is: \[\mathrm{CrBr}_{2}(aq) + \mathrm{Na}_{2}CO_{3}(aq) \longrightarrow \mathrm{CrCO}_{3}(s) + 2\mathrm{NaBr}(aq)\]

- Writing the Products for Reaction d

For NaOH and FeCl3, exchange the ions: Fe pairs with OH to form Fe(OH)3 and Na pairs with Cl to form NaCl. NaCl is soluble, but Fe(OH)3 is insoluble, suggesting a reaction occurs.

- Balancing the Equation for Reaction d

To balance the equation, ensure that the number of atoms of each element on the reactant side is equal to the number of atoms on the product side. The balanced equation is: \[3\mathrm{NaOH}(aq) + \mathrm{FeCl}_{3}(aq) \longrightarrow \mathrm{Fe(OH)}_{3}(s) + 3\mathrm{NaCl}(aq)\]

Key concepts

Double Displacement Reactions

In double displacement reactions, two compounds exchange their ions or elements to form two new compounds. This switcheroo of partners typically occurs when both of the original reactants are in solution. A classic way to visualize this is by imagining two dance partners (cations and anions) swapping to dance with new partners.

For instance, when we mix solutions of LiI and BaS, we're really mixing Li⁺ ions with I^- ions and Ba²⁺ ions with S^2- ions. In a double displacement reaction, Li would pair up with S, and Ba would pair with I to form new compounds. The key to predicting the outcome of such reactions is understanding which new partnerships are viable—this often boils down to the solubility of the potential products in water.

Solubility Rules

The solubility rules are a set of guidelines that help predict whether a compound will dissolve in water. These rules offer general patterns like 'all nitrates are soluble' and 'most sulfates are soluble, except those with Pb²⁺, Ba²⁺, and Ca²⁺'.

In the example reactions, we apply these rules to decide if a product will stay dissolved in water or form a solid precipitate. When we correctly predict the solubility of the products, we can figure out if a reaction will proceed and which compounds will be the end products. It's crucial to remember that soluble compounds will remain in solution and not cause any visible change, while insoluble ones will typically form a precipitate, signaling a chemical reaction has occurred.

Precipitation Reactions

Precipitation reactions are a specific type of double displacement reaction where an insoluble solid, called a precipitate, is formed. From our exercise, the reaction between CrBr₂ and Na₂CO₃ produces chromium(II) carbonate, Cr(CO₃)₂, an insoluble compound that precipitates out of solution. Similarly, the reaction between NaOH and FeCl₃ results in the formation of iron(III) hydroxide, Fe(OH)₃, another insoluble substance.

The identification of these precipitation reactions hinges on the solubility rules. Anything that falls through the 'solubility net' will be a precipitate. Visual cues like cloudiness or a solid forming in the reaction mixture are tell-tale signs of a precipitation reaction taking place.

Balancing Chemical Equations

Balancing chemical equations is quintessential to accurately represent a chemical reaction. It ensures that the number of atoms for each element is the same on both sides of the equation, honoring the Law of Conservation of Mass.

In the precipitate forming reactions (c and d), balancing the equations required ensuring equal numbers of each type of ion on both sides. For Cr(CO₃)₂ to exist as a precipitate, there needs to be a balanced amount of Na⁺ and Br^- ions on the product side, hence the coefficients that precede them in the balanced equation.

• For reaction c, the coefficient 2 in front of NaBr ensures that the number of Br^- ions balances out.
• In reaction d, the coefficient 3 in front of NaOH and NaCl ensures the balance of Na⁺ and OH^- ions.

Mastering the art of balancing equations is not only about the numbers but understanding how matter interacts and transforms in a chemical reaction.