Question

balance the following equation (use the number 1 when needed - dont leave it blank):
ca(no3)2 (aq) + (nh4)3po4 (aq) →
ca3(po4)2 (s) + nh4no3 (aq)

Explanation:

Step1: Balance Ca atoms

We have \(Ca_3(PO_4)_2\) on the product side. So we need \(3\) \(Ca(NO_3)_2\) on the reactant side.
\(3Ca(NO_3)_2(aq)+(NH_4)_3PO_4(aq)\to Ca_3(PO_4)_2(s)+NH_4NO_3(aq)\)

Step2: Balance \(PO_4\) groups

We have \(2\) \(PO_4\) groups in \(Ca_3(PO_4)_2\). So we need \(2\) \((NH_4)_3PO_4\) on the reactant side.
\(3Ca(NO_3)_2(aq)+2(NH_4)_3PO_4(aq)\to Ca_3(PO_4)_2(s)+NH_4NO_3(aq)\)

Step3: Balance \(NH_4\) and \(NO_3\) groups

On the reactant side, from \(3Ca(NO_3)_2\) we have \(6\) \(NO_3\) groups and from \(2(NH_4)_3PO_4\) we have \(6\) \(NH_4\) groups. So we need \(6\) \(NH_4NO_3\) on the product side.
\(3Ca(NO_3)_2(aq)+2(NH_4)_3PO_4(aq)\to Ca_3(PO_4)_2(s)+6NH_4NO_3(aq)\)

Answer:

\(3\), \(2\), \(1\), \(6\)