2AgI+2NH3 give you 2AgNH3 + I2.
Reaction only occurs when ammonia (NH3) is added in excess.
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The equation for reaction between silver nitrate and sodium iodide in water solution is AgNO3 (aq) + NaI (aq) = NaNO3 (aq) + AgI (s).
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.
This is the classic test for halogens. The product will be a precipitate of yellow 'silver iodide'. ( AgI) The reaction eq'n is ZnI2(aq) + 2AgNO3(aq) = 2AgI(s) + Zn(NO3)2(aq)
Silver nitrate + Potassium iodide ----> Silver iodide + Potassium nitrate AgNO3 + KI ----> AgI + KNO3
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There is no reaction, because silver iodide is very insoluble.
Produces Silver iodide precipitate and Sodium nitrate
Silver iodide is not soluble in ammonia solution.
The reaction is:Ag+ + (NO3)- + K + I- = AgI(s) + (NO3)- + K
The equation for reaction between silver nitrate and sodium iodide in water solution is AgNO3 (aq) + NaI (aq) = NaNO3 (aq) + AgI (s).
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.
A yellow precipitate of silver iodiode (AgI) is formed.
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.
Silver(I) iodide
The reaction is the following:AgNO3 + KI = KNO3 = AgI(s)
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.