Question

How many grams of anhydrous remains when 40.0g of KAl(SO4)2 . 12H2O is heated until all water has been liberated?

Answer 1

Upon heating, the water liberates according to the equation:

KAl(SO₄)₂*12H₂O --> KAl(SO₄)₂ + 12H₂O

The molar mass of KAl(SO₄)₂*12H₂O is 474.39 g/mol

The molar mass of anhydrous KAl(SO₄)₂ is 258.21 g/mol

Therefore,

$m(KAl(SO_4)_2)=40.0g(KAl(SO_4)_2\cdot12H_2O)\times\frac{1mol((KAl(SO_4)_2\cdot12H_2O))}{474.39g(KAl(SO_4)_2\cdot12H_2O)}\times\frac{1mol(KAl(SO_4)_2)}{1mol(KAl(SO_4)_2\cdot12H_2O)}\times\frac{258.21g(KAl(SO_4)_2)}{1mol(KAl(SO_4)_2)}=21.8g$

Answer: 21.8 g