What mass of barium sulfate can be produced when 100.0 mL of a .100 M solution of barium chloride is mixed with 100.0 mL of a .100 M solution of iron(iii) sulfate?

Question

anonymous · Answer

To figure this out you need to work out how many moles of barium ions there are before the reaction and whether there are enough sulphate ions to react with them all. If so, then the total will be 0.1M (moles per litre) multiplied by 0.1 litres which gives 0.01 moles. Barium has an atomic weight of 137, oxygen is 16 and sulphur is 32. If you figure out the formula for barium sulphate then you're nearly there.