Question

Complete and balance each equation. If no reaction occurs, write "NO REACTION." a. \(\operatorname{Lil}(a q)+\) BaS \((a q) \longrightarrow\) b. \(\operatorname{KCl}(a q)+\operatorname{CaS}(a q) \longrightarrow\) 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

\(2 \text{LiI}(aq) + \text{BaS}(aq) \longrightarrow \text{BaI}_2(aq) + \text{Li}_2\text{S}(aq)\), NO REACTION, \(\text{CrBr}_2(aq) + \text{Na}_2\text{CO}_3(aq) \longrightarrow \text{CrCO}_3(s) + 2\text{NaBr}(aq)\), \(3\text{NaOH}(aq) + \text{FeCl}_3(aq) \longrightarrow 3\text{NaCl}(aq) + \text{Fe(OH)}_3(s)\).

Step-by-step solution

Predict the products for reaction a

For reaction a, we perform a double replacement reaction. The cations (Li+ and Ba2+) exchange places with the anions (I- and S2-). The potential products are Li2S and BaI2. Since both lithium and barium form soluble salts with these anions, the reaction will take place.

Balance equation a

The balanced equation for reaction a is: \(2 \text{LiI}(aq) + \text{BaS}(aq) \longrightarrow \text{BaI}_2(aq) + \text{Li}_2\text{S}(aq)\). Ensure the number of atoms for each element is the same on both sides of the equation.

Predict the products for reaction b

For reaction b, we would expect a double displacement reaction as K+ and Ca2+ cations swap places with Cl- and S2- anions. However, since all products formed would be soluble in water (KCl, CaS), no precipitate forms and no reaction occurs in aqueous solution.

Record result for reaction b

Since no insoluble products or gases are formed, reaction b is deemed to have 'NO REACTION'.

Predict the products for reaction c

In reaction c, we have a double replacement reaction. The cations (Cr2+ and Na+) and the anions (Br- and CO32-) switch places. The potential products are CrCO3 and NaBr. Since chromium(II) carbonate is insoluble in water, a reaction occurs.

Balance equation c

The balanced chemical equation for reaction c is: \( \text{CrBr}_2(aq) + \text{Na}_2\text{CO}_3(aq) \longrightarrow \text{CrCO}_3(s) + 2\text{NaBr}(aq) \). Balance the number of atoms for each element on both sides of the equation.

Predict the products for reaction d

Reaction d involves a double displacement reaction between NaOH and FeCl3. The products formed would be NaCl and Fe(OH)3. Iron(III) hydroxide is an insoluble compound and thus will precipitate out of solution.

Balance equation d

The balanced equation for reaction d is: \(3\text{NaOH}(aq) + \text{FeCl}_3(aq) \longrightarrow 3\text{NaCl}(aq) + \text{Fe(OH)}_3(s)\). The atoms of each element and charge must be balanced.

Key concepts

Double Replacement Reaction

Double replacement reactions are a type of chemical reaction where the positive ions (cations) and the negative ions (anions) of two different molecules exchange places. This exchange results in the formation of two entirely new compounds. As seen in the textbook exercises, these reactions often occur in solutions where ions can freely move and collide with each other.

In double replacement reactions, we usually observe the interaction of two ionic compounds in an aqueous solution. For instance, when LiI reacts with BaS, Li+ exchanges places with Ba2+, resulting in the formation of new products, BaI2 and Li2S. Students must remember that a reaction only occurs if at least one of the new compounds formed is insoluble, or if a gas or weak electrolyte is produced. If all products are soluble, as with reaction b in the provided exercise, no reaction takes place—labelled as 'NO REACTION'.

Balancing Chemical Equations

Balancing chemical equations is a fundamental practice in chemistry that ensures the law of conservation of mass is observed—meaning the number of atoms of each element is the same on both sides of the equation. When balancing equations, like those in reactions a, c, and d of the provided exercise, start by listing the number of atoms of each element in the reactants and products.

Then, adjust the coefficients (the numbers before compounds) to make sure these numbers are the same on both sides. For example, in the equation for reaction a, the coefficient '2' is placed before LiI to balance the number of Li and I atoms. It's important not to alter the chemical formulas while balancing. The goal is to use the smallest whole number coefficients possible to achieve balance.

Solubility Rules

Understanding solubility rules is essential to predict whether a double replacement reaction will result in a precipitate (an insoluble substance that falls out of solution). These rules are guidelines that suggest the solubility of various compounds when dissolved in water.

For example, most nitrates (NO3-) and alkali metal salts are soluble, while many carbonates (CO32-), phosphates (PO43-), sulfides (S2-), and hydroxides (OH-) are insoluble unless they're paired with an alkali metal or ammonium (NH4+). With reaction c from the exercise, the rule that most carbonates are insoluble unless with an alkali metal or ammonium helps us predict the formation of an insoluble product, CrCO3.

Precipitation Reaction

A precipitation reaction is a kind of double replacement reaction where an insoluble solid called a precipitate is formed from reactants in aqueous solution. This is a key concept for visualizing how soluble ionic compounds can interact to produce an insoluble compound, which is critical for a multitude of applications in scientific and industrial fields.

For instance, in the exercise's reaction d, we mix NaOH with FeCl3, resulting in the formation of Fe(OH)3, an insoluble iron(III) hydroxide precipitate. To identify a precipitation reaction, one can look up the solubility rules or actually observe the formation of a cloudy substance in what was a clear solution, which indicates the formation of a precipitate.