Question

Many plants are poisonous because their stems and leaves contain oxalic acid, H2C2O4, or sodium oxalate, Na2C2O4; when ingested, these substances cause swelling of the respiratory tract and suffocation. A standard analysis for determining the amount of oxalate ion, C2O42-, in a sample is to precipitate this species as calcium oxalate, which is insoluble in water. Write the net ionic equation for the reaction between sodium oxalate and calcium chloride, CaCl2, in aqueous solution. (Use the lowest possible coefficients. Omit states-of-matter in your answer.)

Answer 1

Answer:

H₂C₂O₄ + Ca²⁺ ⟶ CaC₂O₄ + 2H⁺

Explanation:

There are three steps you must follow to get the answer.

You must write the:

1. Molecular equation
2. Ionic equation
3. Net ionic equation

1. Molecular equation

H₂C₂O₄(aq) + CaCl₂(aq) ⟶ CaC₂O₄(s) + 2HCl(aq)

2. Ionic equation

You write the molecular formulas for the weak electrolytes and solids, and you write the soluble ionic substances as ions.

H₂C₂O₄(aq) + Ca²⁺(aq) + 2Cl⁻(aq) ⟶ CaC₂O₄(s) + 2H⁺(aq) + 2Cl⁻(aq)

3. Net ionic equation

To get the net ionic equation, you cancel the ions that appear on each side of the ionic equation.

H₂C₂O₄(aq) + Ca²⁺(aq) + 2Cl⁻(aq) ⟶ CaC₂O₄(s) + 2H⁺(aq) + 2Cl⁻(aq)

The net ionic equation is

H₂C₂O₄ + Ca²⁺ ⟶ CaC₂O₄ + 2H⁺

Answer 2

The net ionic equation is;  Ca^2+ (aq) + C2O4^2-(aq) -------> CaC2O4(s)

First we must put down the molecular equation as follows;

Na2C2O4(aq) + CaCl2(aq) ------> 2NaCl(aq) + CaC2O4(s)

The total ionic equation is;

2Na^+(aq) + Cl^- (aq) +  Ca^2+ (aq) + C2O4^2-(aq) -------> 2Na^+(aq) +  Cl^- (aq) +  CaC2O4(s)

The net ionic equation is;

Ca^2+ (aq) + C2O4^2-(aq) -------> CaC2O4(s)

The spectator ions in this reaction are Na^+ and Cl^-, they appear on both sides of the reaction equation and they do not appear in the final net ionic equation.

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