Question

Write a balanced chemical equation based on the following description:

Answer 1

Answer:

IrBr₃(aq) + 3AgC₂H₃O₂(aq) ⟶ Ir(C₂H₃O₂)₃(aq) + 3AgBr(s)  

Explanation:

Iridium(III) bromide(aq) + silver acetate(aq) ⟶ iridium(III) acetate + silver bromide

You have probably learned NAG SAG and "Cats Cradle Old People," Sally Said.

I've listed the rules below.

$\begin{array}{lll}\textbf{Soluble} & \textbf{Insoluble}\\\textbf{N}\text{itrates} &\textbf{C} \text{arbonates }\\ \textbf{A}\text{cetates} & \textbf{C}\text{hromates}\\ \textbf{G}\text{roup 1} & \textbf{O}\text{ hydrOxides}\\ & \textbf{P} \text{hosphates}\\\end{array}$

$\begin{array}{lll}\textbf{S}\text{ulfates} &\textbf{S} \text{ulfites}\\ \textbf{A}\text{mmonium}& \textbf{S}\text{ulfides}\\& \textbf{P}\text{b halides}\\\textbf{G}\text{roup 17}& \textbf{M}\text{ercury(II) halides}\\&\textbf{S}\text{ilver halides}\\ \end{array}$

The PMS group are the exception to the rule that halides are usually soluble.

We can use the solubility rules to decide which product is the precipitate.

Iridium(III) acetate — soluble (Acetate)

Silver bromide — insoluble (PMS,  Silver halide)

So silver bromide is a precipitate.

The word equation becomes

Iridium(III) bromide(aq) + silver acetate(aq) ⟶ iridium(III) acetate(aq) + silver bromide(s)

In symbols, the equation is

IrBr₃(aq) + 3AgC₂H₃O₂(aq) ⟶ Ir(C₂H₃O₂)₃(aq) + 3AgBr(s)