Question

4. 450g of chromium (III) sulfate reacts with excess potassium phosphate. How many grams of

potassium sulfate will be produced? (ANS: 6.0x10ʻg K2SO4)

Answer 1

`1.6* 103`g of K₂SO₄ will be produced

Step-by-step explanation:

Given
`K_3PO_4` 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),
`K_3PO_4` 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 =
`((522* 450))/(147)`g of K₂SO₄ = 1597.959 g =
`1.59* 103` g =
`1.6* 103` g

So,
`1.6* 103`g of K₂SO₄ will be produced