Answer: 35.72 % of Barium ions will be present in the original unknown compound.
Explanation: The reaction of Barium ions and sodium sulfate is:
Here, Sodium sulfate is present in excess, Barium ions are the limiting reagent because it limits the formation of product.
Now, 1 mole of barium sulfate is produced by 1 mole of Barium ions.
Molar mass of Barium sulfate = 233.38 g/mol
Molar mass of Barium ions = 137.327 g/mol
233.38 g/mol of barium sulfate will be produced by 137.323 g/mol of Barium ions, so
0.4105 grams of barium sulfate will be produced by = of Barium ions
Mass of barium ions = 0.2415 grams
To calculate percentage by mass, we use the formula:
Mass of the solution = 0.6760 grams
Putting the value in above equation, we get
% mass of Barium ions = 35.72%.