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I got this as a question in chemistry so I assumed it was able to form a precipitation reaction.
It is impossible to balance unless you first get the net ionic equation, I then balanced the net ionic equation, I think this is the correct way to do it and I haven't see anyone post anywhere that says otherwise.
FeCl2 (aq)+KOH (aq)--->Fe(OH) (s) +KCl (aq)
This cant be balanced so if you break it down you have
Fe(^2+) + Cl2(^1-) (aq)+K(^1+) + OH(^1-) (aq)--->Fe(OH) (s) +K(^1+) + Cl(^1-) (aq)
You can then cancel out the K+ on both sides and you have
FeCl2+Oh(^-)---.Fe(OH)+Cl
THEN slap a two on the CL on the right side and you have a balanced net ionic equation. This is the only way I found it to work out. I hope this helps and I am 99% sure its correct.
It is a precipitation reaction because you have a solid forming.
I got this as a question in chemistry.
It is impossible to balance unless you first get the net ionic equation, I then balanced the net ionic equation, I think this is the correct way to do it and I haven't see anyone post anywhere that says otherwise.
FeCl2 (aq)+KOH (aq)--->Fe(OH) (s) +KCl (aq)
This cant be balanced so if you break it down you have
Fe(^2+) + Cl2(^1-) (aq)+K(^1+) + OH(^1-) (aq)--->Fe(OH) (s) +K(^1+) + Cl(^1-) (aq)
You can then cancel out the K+ on both sides and you have
FeCl2+Oh(^-)---.Fe(OH)+Cl
THEN slap a two on the CL on the right side and you have a balanced net ionic equation. This is the only way I found it to work out. I hope this helps and I am 99% sure its correct.
There's no such neutral compound as FeCl, as this would imply a +1 charge for the iron ion, which does not exist. However, iron(II) chloride (FeCl2) and iron(III) chloride (FeCl3) are very common in chemistry. Neither of them, however, are precipitates, as chlorides of iron are easily water soluble.
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What is the name of the grey precipitate in the confirmatory test for mercury?
dimeric mercury ion Hg2+ 2 + 2 KI → Hg2I2 + 2 K+2 Hg2+ 2 + 2 NaOH → 2 Hg 2O + 2 Na+ + H2O Confirmation test for mercury:Hg2+ + 2 KI (in excess) → HgI2 + 2 K+HgI2 + 2 KI → K2[HgI4] (red precipitate dissolves)2 Hg2+ + SnCl2 → 2 Hg + SnCl4 (white precipitate turns gray)
How Wagner react to alkaloid?
Dissolve 2.0 grams of iodine and 6.0 grams of KI in 100.0 ml of H2O.
What is the balanced equation of KI plus BaS?
The balanced equation for KI + BaS is 2Kl + BaS -> BaI2 + K2S.
Will BaCl and KI form a reaction?
Yes, when Barium chloride (BaCl2) and Potassium iodide (KI) are mixed, a reaction will occur. BaCl2 and KI will undergo a double displacement reaction to form Barium iodide (BaI2) and Potassium chloride (KCl).
What is the precipitate produced in the bromination reaction?
There need to be more details provided to answer this question. If you are referring to the white precipitate that is produced in the bromination of phenol which is also known as phenylamine by adding bromine(aq) to phenol or phenylamine then the answer is 2,4,6-tribromophenol or it can be written as 2,4,6-tribromophenylamine.
Will a precipitate form if solutions of the soluble salts AgNO3 and KI are mixed?
Yes, a precipitate of AgI (silver iodide) will form because silver iodide is insoluble in water. This reaction can be represented by the equation AgNO3 + KI -> AgI + KNO3.
What cause the precipitate to form when lead nitrate and potassium iodide?
The chemical reaction is:Pb(NO3) 2 + 2 KI = PbI2(s) + 2 KNO3Lead iodide is insoluble in water and form an yellow precipitate.
Will a precipitate form if solutions of the soluble salts HgNO3 and KI are mixed?
Yes, a yellow precipitate of mercury(II) iodide (HgI2) will form when solutions of mercury(II) nitrate (Hg(NO3)2) and potassium iodide (KI) are mixed due to the insolubility of mercury(II) iodide. This reaction is a double displacement reaction where the mercury cation exchanges with the potassium cation to form the insoluble compound.
What is KI plus Cl2?
KI plus Cl2 undergoes a redox reaction to form KCl and I2. The chlorine (Cl2) is reduced to chloride ions (Cl-) and the iodide ions (I-) are oxidized to form elemental iodine (I2).
What is the role of KI in the estimation of aniline?
In the estimation of aniline, KI (potassium iodide) is used as a reagent to form a colored complex with aniline in the presence of an oxidizing agent such as hypochlorite. The intensity of the color formed is directly proportional to the concentration of aniline, allowing for its estimation through spectrophotometric measurement.
What is the name of the grey precipitate in the confirmatory test for mercury?
dimeric mercury ion Hg2+ 2 + 2 KI → Hg2I2 + 2 K+2 Hg2+ 2 + 2 NaOH → 2 Hg 2O + 2 Na+ + H2O Confirmation test for mercury:Hg2+ + 2 KI (in excess) → HgI2 + 2 K+HgI2 + 2 KI → K2[HgI4] (red precipitate dissolves)2 Hg2+ + SnCl2 → 2 Hg + SnCl4 (white precipitate turns gray)
Balance hi plus koh equals ki plus h2o?
The balanced chemical equation for the reaction between HI and KOH is: HI + KOH --> KI + H2O. In this reaction, hydrogen iodide (HI) reacts with potassium hydroxide (KOH) to form potassium iodide (KI) and water (H2O). The equation is balanced in terms of atoms and charge.
Precipitation reactions silver nitrate plus potassium iodide?
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.
How Wagner react to alkaloid?
Dissolve 2.0 grams of iodine and 6.0 grams of KI in 100.0 ml of H2O.
What is the balanced equation of KI plus BaS?
The balanced equation for KI + BaS is 2Kl + BaS -> BaI2 + K2S.
What happen when you add iodide to silver nitrate?
When iodide is added to silver nitrate, a chemical reaction occurs, resulting in the formation of silver iodide precipitate. This can be represented by the equation: AgNO3 + KI -> AgI(s) + KNO3. The silver iodide formed is insoluble in water and appears as a yellow precipitate.
What is the reaction equation for Pb NO3 2 plus KI?
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