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.
Generally not. KOH is water soluble.
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.
bak+si
National Council for Training in the Vocational Trades
Molarity = moles of solute/Liters of solution. get moles KI 2.822 grams KI (1 mole KI/166 grams) = 0.017 moles KI ( 67.94 ml = 0.06794 Liters ) Molarity = 0.017 moles KI/0.06794 Liters = 0.2502 M KI
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.
barium sulfate, BaSO4 The reaction is CuSO4(aq) + Ba(NO3)2(aq) --> Cu(NO3)2(aq) + BaSO4(s)
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..
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 reaction is:Cl2 + 2 Ki = 2 KCl + I2
Dissolve 2.0 grams of iodine and 6.0 grams of KI in 100.0 ml of H2O.
a precipitate. motha nacha -.0
Already balanced
bak+si
Me too i don't know
Pb(CH3COO)2 + 2 KI = PbI2 + 2 K(CH3COO) PbI2 is an yellow precipitate.