PbI₂ (lead iodide) will be formed when Potassium Iodide reacts with Lead (II) Nitrate.
This type of equation which denotes the equal number of atoms of reactants and products in a chemical reaction.
The balanced equation between Potassium Iodide and Lead (II) Nitrate is shown below:
2KI₍s) + Pb(NO₃)₂₍aq) → PbI₂(s) + 2KNO₃(aq)
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