Table of Content
|barium hydroxide||potassium sulfate||potassium hydroxide||barium sulfate|
|(dung dịch)||(rắn)||(dung dịch)||(kt)|
|(không màu)||(trắng)||(không màu)||(trắng)|
|Nguyên - Phân tử khối (g/mol)|
|Khối lượng (g)|
No information found for this chemical equation
Explanation: The ideal environmental conditions for a reaction, such as temperature, pressure, catalysts, and solvent. Catalysts are substances that speed up the pace (velocity) of a chemical reaction without being consumed or becoming part of the end product. Catalysts have no effect on equilibrium situations.
In a full sentence, you can also say Ba(OH)2 (barium hydroxide) reacts with K2SO4 (potassium sulfate) and produce KOH (potassium hydroxide) and BaSO4 (barium sulfate)
This equation does not have any specific information about phenomenon.
In this case, you just need to observe to see if product substance BaSO4 (barium sulfate), appearing at the end of the reaction.
Or if any of the following reactant substances K2SO4 (potassium sulfate), disappearing
We no further information about this chemical reactions.
barium hydroxide2(NH4)3PO4 + 3Ba(OH)2 → 6H2O + 6NH3 + Ba3(PO4)2 2SO2 + Ba(OH)2 → Ba(HSO3)2 2Cl2 + 2Ba(OH)2 → BaCl2 + 2H2O + Ba(ClO)2 View all equations with Ba(OH)2 as reactant
potassium sulfateBa(OH)2 + K2SO4 → 2KOH + BaSO4 MgCl2 + K2SO4 → 2KCl + MgSO4 BaCl2 + K2SO4 → 2KCl + BaSO4 View all equations with K2SO4 as reactant
potassium sulfate2H2O + 2KMnO4 + 5SO2 → 2H2SO4 + 2MnSO4 + K2SO4 2H2O + 2K + CuSO4 → Cu(OH)2 + H2 + K2SO4 10FeO + 18H2SO4 + 2KMnO4 → 5Fe2(SO4)3 + 18H2O + 2MnSO4 + K2SO4 View all equations with K2SO4 as product
potassium sulfate2H2O + 2K + CuSO4 → Cu(OH)2 + H2 + K2SO4 2H2O + 2KMnO4 + 5SO2 → 2H2SO4 + 2MnSO4 + K2SO4 10FeO + 18H2SO4 + 2KMnO4 → 5Fe2(SO4)3 + 18H2O + 2MnSO4 + K2SO4 View all equations with K2SO4 as product