Answer:
1597.959 g  
Explanation:
Given Data:
Amount of Cr₂(SO₄)₃ = 450 g
Amount of potassium phosphate K₃PO₄ = in Excess
grams of potassium sulfate K₂SO₄= ?
Solution 
The Reaction will be 
                  Cr₂(SO₄)₃ + 2K₃PO₄  ----------> 3K₂SO₄ + 2CrPO₄
Information that we have from reaction
                  Cr₂(SO₄)₃ + 2K₃PO₄  ----------> 3K₂SO₄ + 2CrPO₄
                     1 mol          2 mol                   3 mol 
we come to know from the above reaction that
1 mole of chromium(iii) sulfate (Cr₂(SO₄)₃) react with 2 mole of potassium phosphate (K₃PO₄) to produce 3 mole of K₂SO₄
We also know that
molar mass of Cr₂(SO₄)₃ = 147 g/mol
molar mass of K₃PO₄ = 212 g/mol
molar mass of K₂SO₄ = 174 g/mol
if we represent mole in grams then
       Cr₂(SO₄)₃             +       2K₃PO₄          ----------> 3K₂SO₄ + 2CrPO₄
        1 mol (147 g/mol)          2 mol  (212 g/mol)       3 mol  (174g/mol) 
So, Now we have the following details
           Cr₂(SO₄)₃  +  2K₃PO₄          ----------> 3K₂SO₄ + 2CrPO₄
               147 g          424 g                            522 g 
So,
we come to know that 147 g of Cr₂(SO₄)₃ combine with 424 g of 2K₃PO₄ produce  522 g of K₂SO₄
So now we calculate that how many grams of potassium sulfate will be produced 
Apply unity formula
               147 g of  Cr₂(SO₄)₃  ≅ 522 g of K₂SO₄
               450 g of  Cr₂(SO₄)₃  ≅ ? g of K₂SO₄
by doing cross multiplication
g of K₂SO₄ =522 g x 450 g / 147 g
g of K₂SO₄ =  1597.959 g
So the write answer is  1597.959 g  
***Note: By calculation it is obvious that the correct answer is  1597.959 g