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2AgI+2NH3 give you 2AgNH3 + I2.

Reaction only occurs when ammonia (NH3) is added in excess.

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Q: What is reaction between silver iodide and ammonia?
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What is the reaction between silver nitrate and potassium iodide?

learn your chemistry and find out.... Bye


Write word equations to descrbe the reaction silver iodide and potassium acetate?

There is no reaction, because silver iodide is very insoluble.


Reaction of sodium iodide with silver nitrate?

Produces Silver iodide precipitate and Sodium nitrate


What is the chemical equation when NH3 plus AgI?

Silver iodide is not soluble in ammonia solution.


What is the net ionic equation for the reaction that occurs between silver nitrate and potassium iodide?

The reaction is:Ag+ + (NO3)- + K + I- = AgI(s) + (NO3)- + K


Potassium iodide nitric acid?

The equation for reaction between silver nitrate and sodium iodide in water solution is AgNO3 (aq) + NaI (aq) = NaNO3 (aq) + AgI (s).


What are silver salts?

Examples: silver chloride, silver fluoride, silver iodide, silver bromide, silver astatide, silver sulfide, silver nitrate etc. For silver halogenides a method of preparation is the reaction between silver nitrate and a salt containing the halogen.


What is the evidence that a chemical reaction occured in silver nitrate and potassium iodide?

A yellow precipitate of silver iodiode (AgI) is formed.


What is the correct chemical formula for silver iodate?

The molecular formula for silver iodide is AgI.Silver iodide is an inorganic, yellow compound which is used in many things, from silver-based photography to antiseptic.


What is the other term for silver iodide?

Silver(I) iodide


How do you balance silver nitrate plus potassium iodide?

The reaction is the following:AgNO3 + KI = KNO3 = AgI(s)


How much sodium nitrate is produced when 1.7G of silver nitrate and 1.5G of sodium iodide is reacted?

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.