Respuesta :
Answer:
1597.959 g Â
Explanation:
Given Data:
Amount of Cr₂(SO₄)₃ = 450 g
Amount of potassium phosphate K₃PO₄ = in Excess
grams of potassium sulfate Kâ‚‚SOâ‚„= ?
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 Kâ‚‚SOâ‚„ = 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 Kâ‚‚SOâ‚„ =522 g x 450 g / 147 g
g of Kâ‚‚SOâ‚„ = Â 1597.959 g
So the write answer is  1597.959 g Â
***Note: By calculation it is obvious that the correct answer is  1597.959 g Â