A student is given 50.0mL of a solution of  Na2CO3  of unknown concentration. To determine the concentration of the solution, the student mixes the solution with excess 1.0MCa(NO3)2(aq) , causing a precipitate to form. The balanced equation for the reaction is shown below.

Na2CO3(aq)+Ca(NO3)2(aq)→2NaNO3(aq)+CaCO3(s)

(a) Write the net ionic equation for the reaction that occurs when the solutions of  Na2CO3  and  Ca(NO3)2  are mixed.

Respuesta :

Answer:

Ca(aq)⁺²  + CO₃⁻²(aq) → CaCO₃(s)

Explanation:

Breaking down the equation into ionic form gives:

2Na⁺(aq) + CO₃⁻²(aq) + Ca⁺²(aq) + 2NO₃⁻¹ (aq)  →  2Na⁺(aq) + 2NO₃⁻¹(aq)  + CaCO₃(s)

Eliminating all the same ionic states on both sides of the equation gives following final equation

Ca(aq)⁺²  + CO₃⁻²(aq) → CaCO₃(s)

Net ionic equation:-

[tex]Ca^{+2}(aq)+ CO_3^{-2}(aq) \rightarrow CaCO_3(s)[/tex]

Given equation:-

[tex]Na_2CO_3(aq)+Ca(NO_3)_2(aq)\rightarrow2NaNO_3(aq)+CaCO_3(s)[/tex]

Breaking down the equation into ionic form gives:

[tex]2Na^{+} (aq) + CO_3^{-2} (aq) + Ca^{2+}(aq) + 2NO_3^{-}(aq) \rightarrow 2Na^{+} (aq) + 2NO_3^{-} + CaCO_3(s)[/tex]

 Eliminating all the same ionic states on both sides of the equation gives the equation as follows:-

[tex]CO_3^{-2} (aq) + Ca^{2+}(aq) + \rightarrow CaCO_3(s)[/tex]

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