Iron(III)Sulfate and Barium Iodide Reaction:
Fe2(SO4)3 + 3BaI2 → 2FeI3 + 3BaSO4
Fe2(SO4)3 + 3 BaCl2 = 2 FeCl3 + 3 BaSO4
The net ionic equation is Ba^2+ (aq) + SO4 ^2- (aq) --> BaSO 4 (s) in case that helps you.
OH! Try writing (C2H3O2)2 instead of (CH3COO)2
3Ba(NO3)2 + 1Fe2(SO4)3 --> 3BaSO4 + 2Fe(NO3)3
This equation is BaCl2 (aq) + K2SO4 (aq) -> 2 KCl (aq) + BaSO4 (s).
Na2SO4+BaCl2 --> 2NaCl+BaSO4
Ba + FeSO4 --> BaSO4 + Fe
Na2SO4 +CaCl2---------------> 2NaCL +CaSo4
pottasium sulphate and ammonium chloride as a mother liqour
There's nothing to drive the reaction, since all the possible combinations are soluble. Before, you would have copper, sodium, chloride and sulfate ions in solution; after, you would have the same thing. It's not really a reaction if nothing happens.
(NH4)2SO4(aq) + BaCl2(aq) → 2NH4Cl(aq) + BaSO4(s). Barium sulfate is the precipitate.
(NH4)2(SO4) aq + Ba(OH)2 aq ---> Ba(SO4) solid + 2H2O liquid + 2NH3 gas2H2O liquid + 2NH3 gas came from 2NH4OH
BaCl2+K2SO4=2KCl+BaSO4
The BaSO4 (barium sulfate) will precipitate out of solution because it is insoluble, whereas the KCl2 is soluble and will remain dissolved. The balanced equation is: K2SO4 + BaCl2 -----> 2KCl + BaSO4
Sodium chloride and potassium sulfate will not react.
The formula unit for the formation of potassium chloride and barium sulfate is one mole. One unit of potassium sulfate and barium chloride are required for the reaction.
2KOH + H2SO4 -> 2H2O + K2SO4
The balanced equation for sodium sulfate (Na2SO4) plus barium chloride (BaCl2) yielding barium sulfate (BaSO4) and sodium chloride (NaCl) is: Na2SO4 + BaCl2 -> BaSO4 + 2NaCl
BaSO4 + 2CaCl2 --> Ba(Cl2)2 + Ca2SO4
there are no co efficents infront of any of the reactants or products
Na2SO4 +CaCl2---------------> 2NaCL +CaSo4
This equation is BaCl2 (aq) + Na2SO4 (aq) -> 2 NaCl (aq) + BaSO4 (s).
Aluminium Sulphate and potassium hydroxide. Al2(SO4)3 + 6KOH = 3K2(SO4) + 2Al(OH)3
K2SO4(aq) + SrI2(aq)=SrSO4(aq) + 2kI(aq)