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
