Question

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

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.

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)

Answer 2

The balanced chemical reaction for the description: aqueous iridium (III) bromide reacts with aqueous silver acetate to form solid silver bromide and aqueous iridium (III) acetate is $\[IrB{r_3}(aq) + 3C{H_3}COOAg(aq) \to 3AgBr(s) + {(C{H_3}COO)_3}Ir(aq)\]$

Explanation:

• When iridium (III) bromide (aqueous) reacts with silver acetate (aqueous), it results in the formation of aqueous iridium acetate and solid silver bromide.
• The reaction can be written as:

        $\[IrB{r_3}(aq) + C{H_3}COOAg(aq) \to AgBr(s) + {(C{H_3}COO)_3}Ir(aq)\]$

• In a balanced chemical equation, the number of atoms of all the elements on both sides of the reaction must be equal.
• To balance the above reaction, put 3 as coefficient before $\[C{H_3}COOAg\]$ and $\[AgBr\]$.
• Therefore, the balanced chemical equation can be given as:

        $\[IrB{r_3}(aq) + 3C{H_3}COOAg(aq) \to 3AgBr(s) + {(C{H_3}COO)_3}Ir(aq)\]$