Question

Use the equation editor or "Insert Chemistry - WIRIS editor" to write the balanced molecular chemical equation for the reaction of aqueous 0.13 M lead (II) nitrate, with 0.19 M potassium carbonate. You may need to consult Appendix E to determine the states of each reactant and product. Assume any insoluble products are completely insoluble.

Answer 1

Balanced equation:

`Pb(NO_(3))_(2)(aq)+K_(2)CO_(3)(aq)\rightarrow PbCO_(3)(s)+2KNO_(3)(aq)`

Step-by-step explanation:

The chemical reaction between Lead(II) Nitrate and potassium carbonate is as follows.

`Lead(II)Nitrate+Potassium\,carbonate \rightarrow Lead(III)\,\,carbonate+Potassium\,nitrate`

`Pb(NO_(3))_(2)(aq)+K_(2)CO_(3)(aq)\rightarrow PbCO_(3)(s)+2KNO_(3)(aq)`

Ionic equation:

`Pb^(2+)(aq)+2NO_(3)^(-)(aq)+2K^(+)(aq)+CO_(3)^(2-)(aq)\Leftrightarrow PbCO_(3)(s)+K^(+)(aq)+2NO_(3)^(-)`

Cancel the same ions on the both sides of the reaction.

The net ionic equation is as follows.

`Pb^(2+)(aq)+CO_(3)^(2-)(aq)\Leftrightarrow PbCO_(3)(s)`