Question

B) calculate the ratio of moles of h2o to moles of anhydrous kal(so4)2. note: report the ratio to the closest whole number. (show all work including units)

Answer 1

Answer: -

Mass of Hydrated KAl(SO₄)₂ = 2.0 g

Molar mass of anhydrous KAl(SO₄)₂ = 258.20 g/ mol

Mass of of anhydrous KAl(SO₄)₂ = mass of the 2nd heating = A

The mass of water released = mass of the Aluminum Cup + 2.0 grams of KAl(SO₄)₂ - mass of the 2nd heating

= H g

Moles of water released = $\frac{H g}{18 g/mol}$

Moles of anhydrous KAl(SO₄)₂ = $\frac{A g}{258.20 g/mol}$

Required ratio = $\frac{moles of water}{moles of anhydrous KAl(SO4)2}$