KI + CH3COOH --> HI + CH3COO-K+ ROOH + 2HI --> ROH + H2O + I2 TITRATION WITH SODIUM THIOSULPHATE TO DERIVE THE PEROXIDE VALUE 2NA2S2O3 + I2 (PURPLE) --> NA2S4O6 + NaI (COLOURLESS)
There is no real "net ionic" equation, as KI is a catalyst that decomposes H2O2. The "slow step" is H2O2 + I- --> OI- + H2O The "fast step" is H2O2 + OI- --> H2O + I- O2 The overall is: 2 H2O2 --> 2H2O + O2
the required equation is HgCl2+4KI>>2KCl+K2HgI4. according to stoichiometric calculations 4 moles of KI gives 1 mole of k2HgI4 THEREFORE 0.4 moles of K2HgI4 requires----- ? 0.4 moles x 4 moles/1 mole=1.6 moles therefore 1.6 moles of KI is required to produce 0.4 moles of K2HgI4
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.
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
Me too i don't know
Cl2(g) + 2KI --> 2KCl(aq) + I2(s)
Already balanced
H2O2 + 2 KI --> 2 KOH + I2
yhujuhj
The balanced equation is 2 KI + Pb(NO3)2 -> 2 KNO3 + PbI2.
The duration of Thoda Hai Bas Thode Ki Zaroorat Hai is 1440.0 seconds.
Thoda Hai Bas Thode Ki Zaroorat Hai ended on 2010-11-12.
"Thoda Hai Bas Thode Ki Zaroorat Hai" was created in 1999 by Balaji Telefilms.
There is no reaction, because all product ( if reaction goes) would be soluble. so : H+ + Cl- + K+ + I- = H+ + Cl- + K+ + I- and than all of them dare present as a ions only.
KI + NaCl = KCl + NaI
2 KBr + BaI2 ----> 2 KI + BaBr2