Atomic masses are: Ba=137.33, F=19.00, O=16.00, H=1.01.
5(H2O)=5(2.02+16)=5×18.02=90.1; Ba(FO4)2=137.33+2(19+64)=137.33+166=303.33.
Total atomic mass of compound=393.43 of which 303.33 will form the residue.
So the percentage of Ba(FO4)2 in the hydrated barium salt is 303.33/393.43=77.1%.
Therefore 23.4g contains about 18.04g of Ba(FO4)2. This would be the amount of residue after the water had boiled off.