Complete and balance the molecular equation, including phases, for the reaction of aqueous ammonium bromide, NH4Br, and aqueous lead(II) acetate, Pb(C2H3O2).
Enter the balanced net ionic equation, including phases, for this reaction.
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2NH4Br(aq) + Pb(C2H3O2)2(aq) ==> PbBr2(s) + 2NH4C2H3O2(aq)
Pb^2+(aq) + 2Br^-(aq) ==> PbBr2(s)
To complete and balance the molecular equation for the reaction between aqueous ammonium bromide (NH4Br) and aqueous lead(II) acetate (Pb(C2H3O2)), we'll first list the given compounds:
NH4Br (ammonium bromide)
Pb(C2H3O2) (lead(II) acetate)
The molecular equation for this reaction can be written as follows:
NH4Br (aq) + Pb(C2H3O2) (aq) → PBBr2 (s) + 2CH3COOH (aq) + NH4C2H3O2 (aq)
Now let's proceed to balance the equation:
2NH4Br (aq) + Pb(C2H3O2) (aq) → PBBr2 (s) + 2CH3COOH (aq) + 2NH4C2H3O2 (aq)
This is the balanced molecular equation for the reaction between aqueous ammonium bromide and aqueous lead(II) acetate.
To obtain the net ionic equation, we need to eliminate the spectator ions (ions that appear on both sides of the equation without undergoing any changes). In this case, the spectator ions are NH4+ and CH3COO-.
The net ionic equation is as follows:
Pb2+ (aq) + 2Br- (aq) → PbBr2 (s)
So, the balanced net ionic equation for this reaction is:
Pb2+ (aq) + 2Br- (aq) → PbBr2 (s)
To complete and balance the molecular equation, we need to identify the chemical formulas for the products formed when ammonium bromide reacts with lead(II) acetate. Then, we balance the equation by making sure that the number of atoms of each element is the same on both sides of the equation.
First, let's write the chemical formulas for both reactants:
Ammonium bromide: NH4Br
Lead(II) acetate: Pb(C2H3O2)2
Next, let's identify the products formed:
Ammonium and acetate ions will likely remain as dissolved species since both can form soluble salts. However, lead(II) ions will react with bromide ions to form an insoluble solid called lead(II) bromide.
The chemical formula for lead(II) bromide is PbBr2, since lead(II) has a positive 2 charge, and bromide has a negative 1 charge.
Now we can write the balanced equation, including the phases:
NH4Br(aq) + Pb(C2H3O2)2(aq) → PbBr2(s) + (NH4)2(C2H3O2)(aq)
To obtain the net ionic equation, we eliminate the spectator ions (ions that appear on both sides of the equation without undergoing a chemical change). In this case, the acetate ions are spectator ions.
So, the net ionic equation is:
NH4+(aq) + Br-(aq) + Pb2+(aq) → PbBr2(s) + 2NH4+(aq)
Therefore, the balanced net ionic equation, including the phases, for the reaction is:
2NH4+(aq) + 2Br-(aq) + Pb2+(aq) → PbBr2(s) + 2NH4+(aq)