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)
The chemical equation for the reaction between lead nitrate (Pb(NO3)2) and potassium iodide (KI) to form lead iodide (PbI2) and potassium nitrate (KNO3) is: Pb(NO3)2 + 2KI → 2KNO3 + PbI2
Ammonium phosphate and lead nitrate react to form lead phosphate and ammonium nitrate. The balanced chemical equation for this reaction is: (NH4)3PO4 (aq) + 3Pb(NO3)2 (aq) → Pb3(PO4)2 (s) + 3NH4NO3 (aq)
Lead nitrate and potassium bromide react to form lead(II) bromide and potassium nitrate. This chemical reaction is a double displacement reaction where the cations and anions exchange partners to form the new compounds.
The equation cannot be balanced because lead nitrate and lead phosphate contain lead, which has different oxidation states in the two compounds. Balancing redox reactions involving lead compounds is more complex than simple stoichiometry balance.
In the reaction: Lead (Ⅱ) Nitrate + Potassium Iodide → Potassium Nitrate + Lead (Ⅱ) Iodide.. all nitrates are soluble and lead(ii)iodide is insoluble.
The chemical equation for the reaction between lead nitrate (Pb(NO3)2) and potassium iodide (KI) to form lead iodide (PbI2) and potassium nitrate (KNO3) is: Pb(NO3)2 + 2KI → 2KNO3 + PbI2
The chemical reaction between lead nitrate (Pb(NO3)2) and potassium sulfate (K2SO4) produces solid lead sulfate (PbSO4) and potassium nitrate (KNO3) in solution. The balanced chemical equation is: Pb(NO3)2 + K2SO4 -> PbSO4(s) + 2KNO3.
lead nitrate(Pb(NO3)2 + potassium iodide(KI) = lead iodide(PbI) + potassium nitrate (KNO3)
The balanced equation is 2 KI + Pb(NO3)2 -> 2 KNO3 + PbI2.
Ammonium phosphate and lead nitrate react to form lead phosphate and ammonium nitrate. The balanced chemical equation for this reaction is: (NH4)3PO4 (aq) + 3Pb(NO3)2 (aq) → Pb3(PO4)2 (s) + 3NH4NO3 (aq)
When lead (II) nitrate and potassium sulfate are mixed, they will undergo a double displacement reaction. The products of this reaction are lead (II) sulfate and potassium nitrate. This can be represented by the chemical equation: Pb(NO3)2 + K2SO4 -> PbSO4 + 2KNO3.
Equation: Pb(NO3)2 + KI ----> PbI2 + KNO3
It is lead bromide and potassium nitrate
Lead nitrate and potassium bromide react to form lead(II) bromide and potassium nitrate. This chemical reaction is a double displacement reaction where the cations and anions exchange partners to form the new compounds.
The reaction that occurs is a double displacement reaction where lead(II) nitrate and potassium iodide switch partners to form solid lead(II) iodide and potassium nitrate solution. This reaction can be represented by the equation Pb(NO3)2 + 2KI -> Pbl2 + 2KNO3.
When potassium iodide reacts with lead nitrate, a double displacement reaction occurs. The potassium ion and the lead ion switch places to form potassium nitrate and lead iodide. This reaction results in the formation of a yellow precipitate of lead iodide.
The equation cannot be balanced because lead nitrate and lead phosphate contain lead, which has different oxidation states in the two compounds. Balancing redox reactions involving lead compounds is more complex than simple stoichiometry balance.