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Write a balanced chemical equation based on the following description:
aqueous iridium(III) bromide reacts with aqueous silver acetate to form
solid silver bromide and aqueous iridium(III) acetate​

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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.

[tex]\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}[/tex]

[tex]\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}[/tex]

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)

dsdrajlin

The balanced equation between aqueous iridium(III) bromide and aqueous silver acetate to form solid silver bromide and aqueous iridium(III) acetate​ is:

IrBr₃(aq) + 3 AgCH₃COO(aq) ⇒ 3 AgBr(s) + Ir(CH₃COO)₃

Let's consider the unbalanced equation between aqueous iridium(III) bromide and aqueous silver acetate to form solid silver bromide and aqueous iridium(III) acetate​. This is a double displacement equation.

IrBr₃(aq) + AgCH₃COO(aq) ⇒ AgBr(s) + Ir(CH₃COO)₃

We will balance it using the trial and error method. First, we will balance Br atoms by multiplying AgBr by 3.

IrBr₃(aq) + AgCH₃COO(aq) ⇒ 3 AgBr(s) + Ir(CH₃COO)₃

Then, we get the balanced equation by multiplying AgCH₃COO by 3.

IrBr₃(aq) + 3 AgCH₃COO(aq) ⇒ 3 AgBr(s) + Ir(CH₃COO)₃

The balanced equation between aqueous iridium(III) bromide and aqueous silver acetate to form solid silver bromide and aqueous iridium(III) acetate​ is:

IrBr₃(aq) + 3 AgCH₃COO(aq) ⇒ 3 AgBr(s) + Ir(CH₃COO)₃

You can learn more about displacement reactions here: https://brainly.com/question/23008924

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