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.
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)
Dissolve 2.0 grams of iodine and 6.0 grams of KI in 100.0 ml of H2O.
The balanced equation for KI + BaS is 2Kl + BaS -> BaI2 + K2S.
In the reaction between potassium iodide (KI) and silver nitrate (AgNO3), a precipitation reaction occurs, resulting in the formation of silver iodide (AgI), which appears as a solid precipitate, and potassium nitrate (KNO3) in aqueous solution. The balanced equation for the reaction is: 2KI(aq) + AgNO3(aq) → AgI(s) + KNO3(aq). This reaction is commonly used in laboratory settings to demonstrate the formation of insoluble salts.
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).
Yes, AgI is insoluble, therefore will form a precipitate.
The chemical reaction is:Pb(NO3) 2 + 2 KI = PbI2(s) + 2 KNO3Lead iodide is insoluble in water and form an yellow precipitate.
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.
The reaction is:Cl2 + 2 Ki = 2 KCl + I2
ki is added to liberate iodine gas . this liberated iodine gas was then titrated with sodium thiosulphate to give a permanent white precipitate. this white precipitate indicates the endpoint of the titration..
When potassium iodide (KI) reacts with silver nitrate (AgNO₃), a bright yellow precipitate of silver iodide (AgI) is formed. This occurs due to the double displacement reaction between the two compounds. The overall reaction can be represented as: KI + AgNO₃ → AgI (s) + KNO₃. The yellow color of the precipitate is characteristic of silver iodide.
The reaction between potassium iodide (KI) and lead(II) nitrate (Pb(NO₃)₂) results in the formation of lead(II) iodide (PbI₂) as a precipitate and potassium nitrate (KNO₃) in solution. The balanced chemical equation for this double displacement reaction is: 2 KI + Pb(NO₃)₂ → PbI₂ (s) + 2 KNO₃. Lead(II) iodide appears as a bright yellow solid.
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)
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.
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.
Dissolve 2.0 grams of iodine and 6.0 grams of KI in 100.0 ml of H2O.
The balanced equation for KI + BaS is 2Kl + BaS -> BaI2 + K2S.