Do you want that for Monopotassium phosphate, dipotassium phosphate, or tripotassium phosphate? --------------------------------- To clarify for the previous answerer, ionic compounds inherently don't use the mono-, di-, or tri- system used for molecular compounds. Instead, when a cation and an anion is supplied, the ionic compound assumes the number of cations and anions that will generate a neutral ionic compound. In this case, since K is 1+ and PO4 is 3-, the compound potassium phosphate always refers to K3PO4. Therefore: 3AgNO3 + K3PO4 -> 3KNO3 + Ag3PO4
Silver nitrate + Potassium iodide ----> Silver iodide + Potassium nitrate AgNO3 + KI ----> AgI + KNO3
Potassium iodide + silver nitrate --> Silver iodide and potassium nitrate The chemical equation is: K+I- (aq) + Ag+[NO3]- (aq) --> AgI (s) + K+[NO3]- (aq)
Silver nitrate solution and potassium iodide solution can be mixed to form silver iodide due to a double displacement reaction where the silver ions from silver nitrate react with the iodide ions from potassium iodide to form insoluble silver iodide precipitate. The balanced chemical equation for this reaction is: AgNO3 (aq) + KI (aq) → AgI (s) + KNO3 (aq).
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 balanced symbol equation between fluorine and potassium iodide is: 2KI + F2 --> 2KF + I2
The balanced equation is 2 KI + Pb(NO3)2 -> 2 KNO3 + PbI2.
When silver nitrate reacts with potassium iodide, a precipitation reaction occurs where silver iodide is formed. The balanced chemical equation for this reaction is: AgNO3 + KI -> AgI + KNO3. The silver iodide formed will appear as a yellow solid precipitate.
Silver nitrate + Potassium iodide ----> Silver iodide + Potassium nitrate AgNO3 + KI ----> AgI + KNO3
Potassium iodide + silver nitrate --> Silver iodide and potassium nitrate The chemical equation is: K+I- (aq) + Ag+[NO3]- (aq) --> AgI (s) + K+[NO3]- (aq)
lead nitrate(Pb(NO3)2 + potassium iodide(KI) = lead iodide(PbI) + potassium nitrate (KNO3)
When calcium nitrate (Ca(NO₃)₂) reacts with potassium iodide (KI), the balanced chemical equation is: [ \text{Ca(NO}_3\text{)}_2 + 2 \text{KI} \rightarrow \text{CaI}_2 + 2 \text{KNO}_3 ] In this reaction, one mole of calcium nitrate reacts with two moles of potassium iodide to produce one mole of calcium iodide and two moles of potassium nitrate.
A precipitate of Lead iodide and Potassium nitrate are formed
When silver nitrate and potassium iodide are combined, they undergo a double displacement reaction. Silver iodide is formed as a yellow precipitate, while potassium nitrate remains in solution. The balanced chemical equation for this reaction is: AgNO3 + KI -> AgI + KNO3.
Silver nitrate solution and potassium iodide solution can be mixed to form silver iodide due to a double displacement reaction where the silver ions from silver nitrate react with the iodide ions from potassium iodide to form insoluble silver iodide precipitate. The balanced chemical equation for this reaction is: AgNO3 (aq) + KI (aq) → AgI (s) + KNO3 (aq).
A chemical reaction occurs where iron III iodide is formed and potassium nitrate is produced as a byproduct. The balanced chemical equation for this reaction is 2KI + Fe(NO3)3 -> 2KNO3 + FeI3. Iron III iodide is a dark brown solid, whereas potassium nitrate remains in solution.
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
AgNO3(aq) + KI(aq) = KNO3(aq) + AgI(s) This is a classic test for halogens, and AgI precipitates down as a yellow solid.