Question

Partner Namets) For each of the following reactions carried out: Write the balanced chemical equation, the full ionic equation, and the net ionic equation. Beneath the chemical equation, write the name of each reactant and product. Below the net ionic equation, list the identities of any spectator ions. Reaction 1: Pb(NO,)2 (aq) + Kl(a) JUEN Names of compounds: Full Ionic equation: Net Ionic equation spectator ions:

Answer 1

Answer : The full balanced ionic equation will be,

$2KI(aq)+Pb(NO_3)_2(aq)\rightarrow 2KNO_3(aq)+PbI_2(s)$

Reactants are lead nitrate and potassium iodide.

Products are lead iodide and potassium nitrate.

The spectator ions are, $K^+,NO_3^-$

Explanation :

Complete ionic equation : In complete ionic equation, all the substance that are strong electrolyte and present in an aqueous are represented in the form of ions.

Net ionic equation : In the net ionic equations, we are not include the spectator ions in the equations.

Spectator ions : The ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.

When potassium iodide react with lead nitrate then it gives potassium nitrate and lead iodide as a product.

The full balanced ionic equation will be,

$2KI(aq)+Pb(NO_3)_2(aq)\rightarrow 2KNO_3(aq)+PbI_2(s)$

The ionic equation in separated aqueous solution will be,

$2K^+(aq)+2I^{-}(aq)+Pb^{2+}(aq)+2NO_3^{-}(aq)\rightarrow PbI_2(s)+2K^+(aq)+2NO_3^{-}(aq)$

In this equation, $K^+\text{ and }NO_3^-$ are the spectator ions.

By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.

The net ionic equation will be,

$Pb^{2+}(aq)+2I^{-}(aq)\rightarrow PbI_2(s)$