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

Answer:

Balanced equation:

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

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.

$Pb^{2+}(aq)+CO_{3}^{2-}(aq)\Leftrightarrow PbCO_{3}(s)$