Answer:
See explanation.
Explanation:
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In this case, these three equations show how the molecule behaves in terms aqueous and solid species; thus, we first write balanced conventional equation:
[tex]Ca I_2 (aq) + K_2 C_2 O_4 (aq)\rightarrow CaC_2O_4(s)+2KI(aq)[/tex]
Whereas the solid product is CaC₂O₄ based off its low solubility. Next, we ionize the aqueous species to obtain the total ionic equation:
[tex]Ca^{2+} +2I^- + 2K^+ +(C_2 O_4)^{2-} \rightarrow CaC_2O_4(s)+2I^- + 2K^+[/tex]
Finally, we cancel out potassium and iodide ions as they are the spectator ions due to their presence at both reactants and products in order to obtain the net ionic equation:
[tex]Ca^{2+} (aq) +(C_2 O_4)^{2-}(aq) \rightarrow CaC_2O_4(s)[/tex]
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