Which ion is the limiting reagent and which is the reagent in excess when barium sulphate is made? justfy your prediction
Barium sulphate can be made mixing solutions containing the barium and sulphate ions. The net ionic equation for this reaction is
Ba(2+) + SO(2-)4 ----> BaSO4
a) Which ion is the limiting reagent and which is the reagent in excess when barium sulphate is made? justify your prediction
b)why is it important to prepare the barium sulphate mixture as accurately as possible?
This reaction can be done using the acid-base reaction. Meaning the ions involved in the net ionic equation came from the ions of acids and bases. Acid reacts to neutralize bases thus forming salt and water.
In this problem we can have Barium hydroxide (Ba(OH)2) to react with Sulphuric acid (H2SO4).
Ba(OH)2 + H2SO4 ----> BaSO4 + 2H2O
With this kind of reaction, H2SO4 could be the limiting reagent and the Ba(OH)2 is the excess reagent. It is important to keep the mixture as accurate as possible because it will affect the solution of the product. If H2SO4 is in excess, the product is in acidic condition. It will be basic if Ba(OH)2 is in excess.
There are other ways in producing BaSO4.
BaS + H2SO4 ---> BaSO4 + H2S
BaCl2 + Na2SO4 ----> BaSO4 + 2NaCl