39. complete and balance each equation. if no reaction occurs, write no reaction.\na. lii(aq) + bas(aq)…

Explanation:

Step1: Determine products for a

Predict double - displacement products. Cations and anions swap partners. For $\text{LiI}(aq)+\text{BaS}(aq)$, the products would be $\text{Li}_2\text{S}$ and $\text{BaI}_2$. Both $\text{Li}_2\text{S}$ and $\text{BaI}_2$ are soluble (alkali metal sulfides and iodides of most metals are soluble), so no reaction occurs.

Step2: Determine products for b

For $\text{KCl}(aq)+\text{CaS}(aq)$, the double - displacement products are $\text{K}_2\text{S}$ and $\text{CaCl}_2$. Both $\text{K}_2\text{S}$ (alkali metal sulfide) and $\text{CaCl}_2$ are soluble, so no reaction occurs.

Step3: Determine products for c

For $\text{CrBr}_2(aq)+\text{Na}_2\text{CO}_3(aq)$, the double - displacement products are $\text{CrCO}_3$ and $\text{NaBr}$. $\text{CrCO}_3$ is insoluble (most carbonates are insoluble except those of alkali metals and ammonium). The balanced equation is $\text{CrBr}_2(aq)+\text{Na}_2\text{CO}_3(aq)\rightarrow\text{CrCO}_3(s) + 2\text{NaBr}(aq)$.

Step4: Determine products for d

For $\text{NaOH}(aq)+\text{FeCl}_3(aq)$, the double - displacement products are $\text{Fe(OH)}_3$ and $\text{NaCl}$. $\text{Fe(OH)}_3$ is insoluble (most hydroxides are insoluble except those of alkali metals, ammonium, and some alkaline earth metals). The balanced equation is $3\text{NaOH}(aq)+\text{FeCl}_3(aq)\rightarrow\text{Fe(OH)}_3(s)+3\text{NaCl}(aq)$.

Answer:

a. NO REACTION b. NO REACTION c. $\text{CrBr}_2(aq)+\text{Na}_2\text{CO}_3(aq)\rightarrow\text{CrCO}_3(s) + 2\text{NaBr}(aq)$ d. $3\text{NaOH}(aq)+\text{FeCl}_3(aq)\rightarrow\text{Fe(OH)}_3(s)+3\text{NaCl}(aq)$