Question

question 4
1 pts
the net ionic reaction for the balanced equation shown below is:
3 ca(no₃)₂ (aq) + 2 na₃po₄ (aq) → ca₃(po₄)₂ (s) + 6 nano₃ (aq)
2 no₃⁻ (aq) + 2 na⁺ (aq) → nano₃ (s)
3 ca²⁺ (aq) + 6no₃ (aq) + 6 na⁺ (aq) + 2po₄³⁻ (aq) → ca₃(po₄)₂ (s) + 6no₃ (aq) + 6 na⁺ (aq)
3 ca(no₃)₂ (aq) + 2 po₄³⁻ (aq) → ca₃(po₄)₂ (s)
3 ca²⁺ (aq) + 2 po₄³⁻ (aq) → ca₃(po₄)₂ (s)
3 ca²⁺ (aq) + 2 po₄³⁻ (aq) → 6 nano₃ (aq)

Explanation:

Step1: Write the complete ionic equation

First, break all aqueous ionic compounds into ions.
For \(3Ca(NO_3)_2(aq)\), it breaks into \(3Ca^{2 + }(aq)+6NO_3^{-}(aq)\).
For \(2Na_3PO_4(aq)\), it breaks into \(6Na^{+}(aq)+2PO_4^{3 - }(aq)\).
\(Ca_3(PO_4)_2(s)\) remains as it is (insoluble solid).
For \(6NaNO_3(aq)\), it breaks into \(6Na^{+}(aq)+6NO_3^{-}(aq)\).
The complete ionic equation is:
\(3Ca^{2 + }(aq)+6NO_3^{-}(aq)+6Na^{+}(aq)+2PO_4^{3 - }(aq)\to Ca_3(PO_4)_2(s)+6Na^{+}(aq)+6NO_3^{-}(aq)\)

Step2: Cancel out the spectator ions

Spectator ions are \(Na^{+}(aq)\) and \(NO_3^{-}(aq)\) (appear on both sides of the equation).
After canceling them, we get the net - ionic equation:
\(3Ca^{2 + }(aq)+2PO_4^{3 - }(aq)\to Ca_3(PO_4)_2(s)\)

Answer:

\(3Ca^{2 + }(aq)+2PO_4^{3 - }(aq)\to Ca_3(PO_4)_2(s)\) (the fourth option)