From the looks of it this was an experiment done in a lab rather than given by a textbook. If it is then the reaction did not have a 100% yield and, therefore, is the reason why you are not able to get 1.02 grams (your calculations looks right).

"In addition there is more BaCl2 which implies that Na2SO4 is the limiting reagent which it is not."

This statement is false, a sample of substance that is heavier than another does not always mean it has more molecules. You have shown that there are equal number of molecules of BaCl2 and Na2SO4, 0.00073 mol. When finding the limiting reagent, it is the number of molecules that matters not the weight of the substance. Looking at the reaction, BaCl2+Na2SO4---->BaSO4+2NaCl. For every mol of BaCl2, one mol of Na2SO4 is used. If you have equal numbers of molecules of BaCl2 and Na2SO4, then there is no limiting reagent.

If you want to find the limiting reagent you need to know how many moles of each reactant there is. Let's have a look at this example:

BaCl2+Na2SO4---->BaSO4+2NaCl

There is 1.0g of BaCl2 and 1.0g of Na2SO4, which is the limiting reagent?

"First convert grams into moles"

1.0g BaCl2 * (1 mol BaCl2 / 208.2g BaCl2) = 4.8 x 10^-3 mol BaCl2

1.0g Na2SO4 * (1 mol Na2SO4 / 142.04g Na2SO4) = 7.0 x 10^-3 mol Na2SO4

"Find the ratio of molecules of the reactants"

(7.0 x 10^-3 mol Na2SO4 / 4.8 x 10^-3 mol BaCl2 ) = 1.5 mol Na2SO4 / mol BaCl2

"From this ratio compare it to the equation, BaCl2+Na2SO4---->BaSO4+2NaCl"

The equation shows that for every mol of BaCl2 requires 1 mol of Na2SO4. But we found that there is 1.5 mol of Na2SO4 per mol of BaCl2. Therefore, BaCl2 is the limiting reagent.