Question

Lead (ll) nitride+ammonium sulfate > lead(ll) sulfate + ammonium nitride

Answer 1

Answer:

$Pb_3N_2 + 3 (NH_4)_2SO_4\rightarrow 3 PbSO_4 + 2 (NH_4)_3N$

Explanation:

Let's rewrite the given word equation in its chemical balanced equation representation:

1. Lead(II) nitride is represented by lead, Pb, in an oxidation state of 2+, while nitride is a typical nitrogen anion with a state 3-. As a result, the lowest common multiple between 2 and 3 is 6, meaning 2 lead cations are needed to balance 3 nitrogen anions: $Pb_3N_2$.

2. Ammonium sulfate consists of an ammonium cation with a 1+ charge and sulfate anion with a 2- charge, two ammonium cations needed: $(NH_4)_2SO_4$.

3. Lead(II) sulfate would have one lead cation and one sulfate anion, as they have the same magnitude of charges with opposite signs: $PbSO_4$.

4. Ammonium nitride would require three amonium cations to balance the nitride anion: $(NH_4)_3N$.

Let's write the balanced equation:

$Pb_3N_2 + 3 (NH_4)_2SO_4\rightarrow 3 PbSO_4 + 2 (NH_4)_3N$