Question

question 3
1 pts
the complete ionic reaction for the balanced equation shown below is:
3 ca(no₃)₂ (aq) + 2 na₃po₄ (aq) → ca₃(po₄)₂ (s) + 6 nano₃ (aq)
3 ca²⁺ (aq) + 6no₃ (aq) + 6 na⁺ (aq) + 2po₄³⁻ (aq) → ca₃(po₄)₂ (s) + 6no₃ (aq) +
6 na⁺ (aq)
3 ca²⁺ (aq) + 2 po₄³⁻ (aq) → 6 nano₃ (aq)
2 no₃⁻ (aq) + 2 na⁺ (aq) → nano₃ (s)
3 ca²⁺ (aq) + 2 po₄³⁻ (aq) → ca₃(po₄)₂ (aq)
3 ca(no₃)₂ (aq) + 2 po₄³⁻ (aq) → ca₃(po₄)₂ (s)

Explanation:

Step1: Write the dissociation of reactants

For \(Ca(NO_3)_2\) in aqueous solution: \(Ca(NO_3)_2(aq)=Ca^{2 +}(aq)+2NO_3^{-}(aq)\). So \(3Ca(NO_3)_2(aq)\) dissociates to \(3Ca^{2 +}(aq)+6NO_3^{-}(aq)\).
For \(Na_3PO_4\) in aqueous solution: \(Na_3PO_4(aq)=3Na^{+}(aq)+PO_4^{3 -}(aq)\). So \(2Na_3PO_4(aq)\) dissociates to \(6Na^{+}(aq)+2PO_4^{3 -}(aq)\).

Step2: Write the dissociation of products

For \(NaNO_3\) in aqueous solution: \(NaNO_3(aq)=Na^{+}(aq)+NO_3^{-}(aq)\). So \(6NaNO_3(aq)\) dissociates to \(6Na^{+}(aq)+6NO_3^{-}(aq)\). \(Ca_3(PO_4)_2\) is a solid and does not dissociate.

Step3: Combine the dissociated ions

The complete ionic equation is formed by writing all the ions from reactants and products (except the solid). So \(3Ca^{2 +}(aq)+6NO_3^{-}(aq)+6Na^{+}(aq)+2PO_4^{3 -}(aq)\to Ca_3(PO_4)_2(s)+6NO_3^{-}(aq)+6Na^{+}(aq)\)

Answer:

\(3Ca^{2 +}(aq)+6NO_3^{-}(aq)+6Na^{+}(aq)+2PO_4^{3 -}(aq)\to Ca_3(PO_4)_2(s)+6NO_3^{-}(aq)+6Na^{+}(aq)\) (the first option)