2Ag(aq)+ CrO4(s)---->AgCrO4(s)
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)
the chemical reaction between silver nitrate and potassium chromate in generally used in a titration to look for chloride ions , and is a two step reaction: i will give you the ionic equations, which is pretty much all you need Ag+(aq) + Cl-(s) --> AgCl(s) this part of the equation caused the solution to go cloudy. when all the chloride ions are used up then the silver reacts with the chromate ions to produce the red colour you see when the end point of the precipitation is reached: 2Ag+(aq) + CrO4 2-(aq) --> Ag2 CrO4(s) which produces the red colour the amount of silver nitrate relates directly to the chloride ion concentration as it is a 1:1 ration reaction. i hope this answers your question =)
Since silver chromate has a 1:1 molar ratio with silver nitrate, 4 moles of silver nitrate will produce 4 moles of silver chromate.
In this reaction, the precipitate formed would be silver chromate due to the double displacement reaction between sodium chromate (Na2CrO4) and silver nitrate (AgNO3). Silver chromate is insoluble in water, so it will precipitate out of the solution as a solid, appearing as a yellow precipitate.
The number of moles of silver chromate formed will depend on the stoichiometry of the reaction between silver nitrate and potassium chromate. You need to know the balanced chemical equation, as well as the exact volumes and concentrations of the silver nitrate and potassium chromate solutions to calculate the number of moles of silver chromate formed.
The ionic equation for the reaction between silver nitrate (AgNO3) and potassium chromate (K2CrO4) is: Ag+ + 2NO3- + 2K+ + CrO42- -> Ag2CrO4(s) + 2KNO3 This reaction forms silver chromate (Ag2CrO4) as a solid precipitate, with potassium nitrate (KNO3) remaining in solution.
To find the limiting reactant, calculate the moles of silver nitrate and potassium chromate. Convert the limiting reactant to moles of silver chromate using the balanced chemical equation. Here, 2 moles of silver nitrate react with 1 mole of potassium chromate to form 2 moles of silver chromate. Calculate the moles of silver chromate that can be formed based on the limiting reactant.
Potassium nitrate is too stable and so is silver for these two species to react. There is thus no balanced equation.
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 chromate is not soluble in water.
the chemical reaction between silver nitrate and potassium chromate in generally used in a titration to look for chloride ions , and is a two step reaction: i will give you the ionic equations, which is pretty much all you need Ag+(aq) + Cl-(s) --> AgCl(s) this part of the equation caused the solution to go cloudy. when all the chloride ions are used up then the silver reacts with the chromate ions to produce the red colour you see when the end point of the precipitation is reached: 2Ag+(aq) + CrO4 2-(aq) --> Ag2 CrO4(s) which produces the red colour the amount of silver nitrate relates directly to the chloride ion concentration as it is a 1:1 ration reaction. i hope this answers your question =)
Since silver chromate has a 1:1 molar ratio with silver nitrate, 4 moles of silver nitrate will produce 4 moles of silver chromate.
AgNO3(aq) + KI(aq) = KNO3(aq) + AgI(s) This is a classic test for halogens, and AgI precipitates down as a yellow solid.
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.
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