Question #198727

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

Expert's answer

Upon heating, the water liberates according to the equation:

KAl(SO_{4})_{2}*12H_{2}O --> KAl(SO_{4})_{2} + 12H_{2}O

The molar mass of KAl(SO_{4})_{2}*12H_{2}O is 474.39 g/mol

The molar mass of anhydrous KAl(SO_{4})_{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

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