Produces Silver iodide precipitate and Sodium nitrate
This is a precipitation reaction. Halides of silver are insoluble in water (except silver fluoride) whereas all nitrates are soluble in water. Sodium salts are soluble. Thus, silver iodide is the precipitate. Formula: AgNO3(aq) + NaI(aq) -> AgI(s) + NaNO3(aq)
The chemical equation is: Na+I- (aq) + Ag+[NO3]- (aq) --> AgI (s) + Na+[NO3]- (aq)
A white precipitate of silver iodide forms due to the reaction between silver ions and iodide ions, leaving potassium nitrate in solution. This reaction is a double displacement reaction and is used as a test for iodide ions.
Equation: NaI + AgNO3 ----> NaNO3 + AgI
When reactants lead(II) nitrate and sodium iodide are combined, a double displacement reaction occurs. Lead(II) iodide (insoluble in water) and sodium nitrate are formed, leading to a white precipitate of lead(II) iodide and a solution of sodium nitrate.
This is a precipitation reaction. Halides of silver are insoluble in water (except silver fluoride) whereas all nitrates are soluble in water. Sodium salts are soluble. Thus, silver iodide is the precipitate. Formula: AgNO3(aq) + NaI(aq) -> AgI(s) + NaNO3(aq)
The chemical equation is: Na+I- (aq) + Ag+[NO3]- (aq) --> AgI (s) + Na+[NO3]- (aq)
A white precipitate of silver iodide forms due to the reaction between silver ions and iodide ions, leaving potassium nitrate in solution. This reaction is a double displacement reaction and is used as a test for iodide ions.
The reaction between lead(II) nitrate and sodium iodide is a double replacement reaction. The products of this reaction are lead(II) iodide and sodium nitrate, formed through the exchange of ions.
The equation for the reaction between silver nitrate and sodium iodide is AgNO3 + NaI -> AgBr + NaNO3. The gram formula masses are 169.87 for silver nitrate, 149.89 for sodium iodide, and 84.99 for sodium nitrate. Therefore, 1.7 g of silver nitrate constitutes 1.7/169.87 or 0.010 formula mass of silver nitrate and 1.5 g of sodium iodide constitutes 1.5/149.89 or 0.010 mole of sodium iodide, to the justified number of significant digits. The reaction equation shows that the number of formula unit masses of each reactant and product are the same, so that there will be 0.85 g of sodium nitrate produced, to the justified number of significant digits.
When lead nitrate is mixed with sodium iodide, a solid precipitate of lead iodide is formed along with sodium nitrate. This reaction is a double displacement reaction where the cations of the two compounds switch partners to form the products. Lead iodide is a yellow precipitate that can be easily observed in the reaction mixture.
When aqueous solutions of silver nitrate and sodium iodide [note correct spelling] are mixed, silver iodide solid precipitates from the mixture.
The net ionic equation for the reaction between sodium iodide (NaI) and silver nitrate (AgNO3) when a precipitate is formed is: 2Ag+ + 2I- -> Ag2I (s) This equation represents the formation of silver iodide (AgI) precipitate when silver cations react with iodide anions.
Lead(II) nitrate and sodium iodide will yield lead(II) iodide and sodium nitrate. This is a double displacement reaction, where the cations and anions switch partners resulting in the formation of two new compounds.
When sodium nitrate (NaNO3) reacts with potassium iodide (KI), it forms sodium iodide (NaI) and potassium nitrate (KNO3). This reaction is a double replacement reaction, where the positive ions from each compound switch places. The chemical equation for this reaction is: NaNO3 + KI → NaI + KNO3.
Equation: NaI + AgNO3 ----> NaNO3 + AgI
When reactants lead(II) nitrate and sodium iodide are combined, a double displacement reaction occurs. Lead(II) iodide (insoluble in water) and sodium nitrate are formed, leading to a white precipitate of lead(II) iodide and a solution of sodium nitrate.