250 ml of 0.5M sodium asulphate solution are added to an aq. solution containing 10.0 g of barium chloride resulting in the formation of white precipitate of BaSO4.How many moles and how many grams of barium sulphate will be obtained? At. mass-Ba=137U Cl=35.5U S=32U O=16U.
1 vol 1 vol 1 vol 2 vol
1 mol 1 mol 1 mol
208 g 142 g 233 g
No. of moles of BaCl2 = 10/208 g = 0.048 mol
Since molarity is no. of moles per litre or no. of moles present in 1000 ml of solution
So, 1000 ml of Na2SO4 solution contains = 0.5 mol
Hence, 250 ml of Na2SO4 solution will contain = 0.5x250/1000 = 0.125 mol
Since 1 mol of barium chloride reacts with 1 mol of sodium sulphate so, 0.048 mol of BaCl2 will react with 0.048 mol of Na2SO4 only.
Now, 1 mol of BaCl2 gives = 1 mol of BaSO4
So, 0.048 mol of BaCl2 will give = 0.048 mol of BaSO4
Grams of BaSO4 produced will be = no. of moles x molar mass = 0.048 x 233 = 11.1 g