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)

Moles are defined as the base unit of an element, such that 1 mole is equivalent to 6.022 particles. In the given question, ratio of moles of HO to the moles of anhydrous KAl(SO) is 12:1.

Given that,

Now,

Now, the ratio of HO to KAl(SO) is:

The 11.7 to its closest whole number is 12.

Therefore, the required ratio between water and aluminum potassium sulphate is 12:1.

To know more about moles, refer to the following link:

link

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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 =

Moles of anhydrous KAl(SO₄)₂ =

Required ratio =

the answer is the first option

explanation:

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