There are two possible products when solutions of:
...lead nitrate, Pb(NO3)2, and potassium phosphate, K3PO4, are combined.
Step 1: To determine the products "swap" the cations:
...potassium nitrate ...and...lead phosphate
Step 2: Write the balanced "molecular" equation:
3 Pb(NO3)2 + 2 K3PO4Pb3(PO4)2 + 6 KNO3
Step 3: Classify each of the substances as Soluble (sol) or Not Soluble (ns):
3Pb(NO3)2+2K3PO4Pb3(PO4)2+6KNO3solsolnssol
Step 4: Dissociate all soluble salts, and strong acids. Leave together all "not soluble" salts and weak acids or bases:
3 Pb2+ (aq) + 6 NO3- (aq) + 6K+ (aq) + 2 PO43- (aq) Pb3(PO4)2 (s) + 6 K+ (aq) + 6 NO3- (aq)
Step 5: Cross out "spectator ions", ones that appear on both sides of the reaction (these ions do not participate in the chemistry) and rewrite the "net" reaction using the smallest possible coefficients.
3 Pb2+(aq) + 2PO43-(aq) Pb3(PO4)2 (s)
There's no reaction, all those salts (ammonium AND potassium) are soluble.
lead nitrate + potassium bromide --> lead bromide + potassium nitrate
In the reaction: Lead (Ⅱ) Nitrate + Potassium Iodide → Potassium Nitrate + Lead (Ⅱ) Iodide.. all nitrates are soluble and lead(ii)iodide is insoluble.
Yes, of course: 4 (NH4)3PO4 + 3 Pb(NO3)4 -> Pb3(PO4)4 + 12 NH4NO3 for lead (IV) nitrate and 2 (NH4)3PO4 + 3 Pb(NO3)2 -> Pb3(PO4)4 + 6 NH4NO3 for lead (II) nitrate.
2KI+Pb(NO(3))(2) yields 2KNO(3)+PbI(2). You basically get potassium nitrate and lead (II) iodide when you react potassium iodide and lead nitrate dissolved in solution.
Lead Phosphate and Ammonium Nitrate. 4 (NH4)3PO4 + 3Pb(NO3)4 ----> Pb3(PO4)4 + 12 NH4NO3
lead nitrate + potassium bromide --> lead bromide + potassium nitrate
Lead nitrate + potassium sulfate ---> Lead sulfate + Potassium nitrate
It is lead bromide and potassium nitrate
It produces Potassium nitrate and Lead iodide
The balanced equation is 2 KI + Pb(NO3)2 -> 2 KNO3 + PbI2.
In the reaction: Lead (Ⅱ) Nitrate + Potassium Iodide → Potassium Nitrate + Lead (Ⅱ) Iodide.. all nitrates are soluble and lead(ii)iodide is insoluble.
A precipitate of Lead iodide and Potassium nitrate are formed
lead nitrate(Pb(NO3)2 + potassium iodide(KI) = lead iodide(PbI) + potassium nitrate (KNO3)
Potassium Iodide- KI Lead Nitrate- Pb(NO3)2
Yes, of course: 4 (NH4)3PO4 + 3 Pb(NO3)4 -> Pb3(PO4)4 + 12 NH4NO3 for lead (IV) nitrate and 2 (NH4)3PO4 + 3 Pb(NO3)2 -> Pb3(PO4)4 + 6 NH4NO3 for lead (II) nitrate.
Equation: Pb(NO3)2 + KI ----> PbI2 + KNO3
Yes. The equation is Pb(NO3)2(aq) + 2KI(aq) ==> PbI2(s) + 2 KNO3(aq)