Question

Aqueous Copper (II) nitrate reacts with aqueous potassium iodide to form Copper (II) iodide solid and potassium nitrate

Answer 1

When aqueous Copper (II) nitrate reacts with aqueous potassium iodide, a double-replacement reaction occurs, resulting in the formation of Copper (II) iodide solid and potassium nitrate in solution.

Step-by-step explanation:

When aqueous Copper (II) nitrate reacts with aqueous potassium iodide, a double-replacement reaction occurs. The copper (II) cation (Cu²⁺) from the copper (II) nitrate combines with the iodide anion (I⁻) from the potassium iodide to form Copper (II) iodide (CuI) as a solid precipitate. The potassium cation (K⁺) from the potassium iodide combines with the nitrate anion (NO₃⁻) from the copper (II) nitrate to form potassium nitrate (KNO₃) in solution.

Answer 2

Cu(NO₃)₂ (aq) + 2 KI (aq) ---> CuI₂ (s) + 2 KNO₃

Step-by-step explanation:

When writing the reaction with the symbols, you need to take into account the charges of the ions. If he charges on the ions do not balance out in a molecule, they need to be made up for in the form of subscripts. For example, copper (+2) and iodine (-1) have charges which do not balance. Thus, to make the molecule neutral, you need to have two iodine atoms (CuI₂).

The unbalanced equation:

Cu(NO₃)₂ (aq) + KI (aq) ---> CuI₂ (s) + KNO₃

Reactants: 1 copper, 2 nitrate, 1 potassium, 1 iodine

Products: 1 copper, 1 nitrate, 1 potassium, 2 iodine

The balanced equation:

1 Cu(NO₃)₂ (aq) + 2 KI (aq) ---> 1 CuI₂ (s) + 2 KNO₃

Reactants: 1 copper, 2 nitrate, 2 potassium, 2 iodine

Products: 1 copper, 2 nitrate, 2 potassium, 2 iodine