4. 450g of chromium (iii) sulfate reacts with excess potassium phosphate. how many grams of
potassium sulfate will be produced? (ans: 6.0x10ʻg k2so4)

g of K₂SO₄ will be produced

Explanation:

Given is available in excess

Reaction:

Cr₂(SO₄)₃ + 2K₃PO₄  ----------> 3K₂SO₄ + 2CrPO₄

From the above reaction, it is clear that the moles of (Cr₂(SO₄)₃), and  K₂SO₄  are 1, 2 and 3 respectively

We can say that 1 mole of chromium(iii) sulfate (Cr₂(SO₄)₃) react with 2 mole of potassium phosphate (K₃PO₄) to produce 3 mole of K₂SO₄

molar mass of Cr₂(SO₄)₃ = 147 g/mol

molar mass of K₃PO₄ = 212 g/mol

molar mass of K₂SO₄ = 174 g/mol

We can write as;

     Cr₂(SO₄)₃               +       2K₃PO₄             ----------> 3K₂SO₄  + 2CrPO₄

     1 mol (147 g/mol)         2 mol  (212 g/mol)           3 mol  (174g/mol)

Therefore, we have

         Cr₂(SO₄)₃  +  2K₃PO₄          ----------> 3K₂SO₄ + 2CrPO₄

             147 g         424 g                           522 g

So, we can see that 147 g of Cr₂(SO₄)₃ reacts with 424 g of 2K₃PO₄ to produce  522 g of K₂SO₄

             147 g of  Cr₂(SO₄)₃  = 522 g of K₂SO₄

             So, 450 g of  Cr₂(SO₄)₃ = g of K₂SO₄ = 1597.959 g = g = g

So,g of K₂SO₄ will be produced