How many oxygen atoms are contained in 2.74g of Al2(SO4)3?

Answer 1

= 2.74 grams Al2(SO4)3 x 1 mol Al2(SO4)3 / 342.17 grams Al2(SO4)3 x
12 mol O / 1 mol Al2(SO4)3 x 6.022×10^23 Oxygen atoms / 1mole O
= 5.79 x 10^22 Oxygen atoms

Answer 2

Moles aluminium sulphate = 2.74 g / 342.15 = 0.0080
Moles O = 12 x 0.0080=0.096
mass O = 16 x 0.096=1.54 g

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