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.

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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 =

c is having 4 valence electrons and is bonded to three hydrogens and one sulphur atom, therefore, the number of bond pair is equal to four. as all the valence electrons are used in forming the bonds so, there is no lone pair left on carbon.

s is having 6 valence electrons and is bonded to one other sulfur atom and one carbon atom, therefore, the number of bond pair is equal to 2. as only two valence electrons are used in forming the bonds so, four electrons are left. therefore, two lone pairs are present on s.

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