Question

1. __ (nh₄)₃po₄ + pb(no₃)₄ → pb₃(po₄)₄ + __ nh₄no₃

Explanation:

Step1: Balance \(Pb\) atoms

On the right - hand side, there are \(3\) \(Pb\) atoms in \(Pb_{3}(PO_{4})_{4}\). So, we put a coefficient of \(3\) in front of \(Pb(NO_{3})_{4}\) on the left - hand side.
\((NH_{4})_{3}PO_{4}+3Pb(NO_{3})_{4}\to Pb_{3}(PO_{4})_{4}+NH_{4}NO_{3}\)

Step2: Balance \(PO_{4}\) groups

On the right - hand side, there are \(4\) \(PO_{4}\) groups in \(Pb_{3}(PO_{4})_{4}\). So, we put a coefficient of \(4\) in front of \((NH_{4})_{3}PO_{4}\) on the left - hand side.
\(4(NH_{4})_{3}PO_{4}+3Pb(NO_{3})_{4}\to Pb_{3}(PO_{4})_{4}+NH_{4}NO_{3}\)

Step3: Balance \(NH_{4}\) and \(NO_{3}\) groups

On the left - hand side, from \(4(NH_{4})_{3}PO_{4}\), there are \(4\times3 = 12\) \(NH_{4}\) groups, and from \(3Pb(NO_{3})_{4}\), there are \(3\times4=12\) \(NO_{3}\) groups.
So, we put a coefficient of \(12\) in front of \(NH_{4}NO_{3}\) on the right - hand side.
\(4(NH_{4})_{3}PO_{4}+3Pb(NO_{3})_{4}\to Pb_{3}(PO_{4})_{4}+12NH_{4}NO_{3}\)

Answer:

\(4(NH_{4})_{3}PO_{4}+3Pb(NO_{3})_{4}\to Pb_{3}(PO_{4})_{4}+12NH_{4}NO_{3}\)