Answer:
See explanation.
Step-by-step 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:
![Ca I_2 (aq) + K_2 C_2 O_4 (aq)\rightarrow CaC_2O_4(s)+2KI(aq)](https://img.qammunity.org/2022/formulas/chemistry/college/7qg4izuthi22ti1yqx83g71nisj7gbh09e.png)
Whereas the solid product is CaC₂O₄ based off its low solubility. Next, we ionize the aqueous species to obtain the total ionic equation:
![Ca^(2+) +2I^- + 2K^+ +(C_2 O_4)^(2-) \rightarrow CaC_2O_4(s)+2I^- + 2K^+](https://img.qammunity.org/2022/formulas/chemistry/college/a0ceycethcuz8j4d4hsdhlsqscaxgbktoa.png)
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:
![Ca^(2+) (aq) +(C_2 O_4)^(2-)(aq) \rightarrow CaC_2O_4(s)](https://img.qammunity.org/2022/formulas/chemistry/college/dp5skzmsvubcovqyefayoydx7zucma1hhe.png)
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