White precipitate will formed which is barium sulfate.
A white precipitate. This is the classic test for sulphates.
2Al(OH)3 + 3H2SO4 -> Al2(SO4)3 + 6H2O Looks like aluminum hydroxide and sulfuric acid will form the salt aluminum sulfate and water.
The barium ion in barium hydroxide and sulfate ion in sulfuric acid combine to form barium sulfate, which is insoluble in water.
Ba(OH)2 + H2SO4 react to form BaSO4 + 2H2O
sulphuric acid
Platinum does not react with sulfuric acid under normal conditions.
2Al(OH)3 + 3H2SO4 -> Al2(SO4)3 + 6H2O Looks like aluminum hydroxide and sulfuric acid will form the salt aluminum sulfate and water.
The barium ion in barium hydroxide and sulfate ion in sulfuric acid combine to form barium sulfate, which is insoluble in water.
Barium sulfate is the precipitate
Ba(OH)2 + H2SO4 ---> BaSO4 + 2H2O So one molecule of sulphuric acid will be neutralized by one molecule of barium hydroxide. Or two molecules of sulphuric acid will be neutralized by two molecules of barium hydroxide.
This base is zinc hydroxide.
In the acid-base reaction where sodium hydroxide and sulfuric acid react, the formula is: H2SO4 + 2NaOH --> Na2SO4 + 2H2O. The coefficients shown are necessary to uphold the law of conservation of mass. So, if you have 17 moles of sulfuric acid, you will need twice as many moles of sodium hydroxide, so the answer is 34 moles NaOH.
Ba(OH)2 + H2SO4 react to form BaSO4 + 2H2O
sulphuric acid
Platinum does not react with sulfuric acid under normal conditions.
Barium fluoride
Mix equal volumes of equimolar solutions of sulfuric acid and barium hydroxide. What you will get is a white precipitate of barium sulfate and water (and it will get REALLY hot because of the exothermic formation of water from H+ and OH-).
BaSO4