5.34g of salt of formula M2
(where M is a metal) were dissolved in water. The sulphate ion was precipitated by adding excess barium chloride solution when 4.66g of barium sulphate (BaSO4
) were obtained.
a) How many moles of sulphate ion were precipitated as barium sulphate?
b) How many moles of M2
were in solution?
a) Barium and sulphate exist in a one-to-one ratio in the compound , so the number of moles of barium sulphate equals the number of moles of sulphate. Beside, how to get the exact value of moles?