Respuesta :

zelinsky
First, it is necessary to have the molar mass of FeSO4 • 7H2O and molar mass of compound without water FeSO4.

M(FeSO4 • 7H2O) = 54+32+(16x4)+7(2+16) = 276 g/mole

M(FeSO4) = 54+32+(16x4) = 150 g/mole

When the molar mass value of iron(II) sulfate heptahydrate is subtracted from the molar mass value of iron(II) sulfate we get the total molar mass of water in crystal:

M(H20) = M(FeSO4 • 7H2O) - M(FeSO4) = 276 - 150 = 126 g mole 

Now we can determine how much water there are in 3.2 g of the crystal:

M(FeSO4 • 7H2O) : M(H20) = m(FeSO4 • 7H2O) : m(H20) 

276 : 126 = 3.38 : x

x = 425.88 / 276 = 1.54 g of H2O 

mickymike92

The mass due to water of crystallization present in a 3.38 g sample of FeSO4 •7H2O is 1.53 g

What is water of crystallization?

The water of crystallization is the moles of water present in one molecule of a salt which forms part of the crystal lattice.

For the given salt:

molar mass of salt FeSO4 •7H2O = 278 g/mol

mass of water on 1 mole FeSO4 •7H2O = 7 × 18 = 126 g

Mass due to waters of crystallization present in a 3.38 g sample of FeSO4 •7H2O = 3.38 × 126/278

Mass of water of crystallization = 1.53 g

Therefore, the mass due to water of crystallization present in a 3.38 g sample of FeSO4 •7H2O is 1.53 g.

Learn more about water of crystallization at: https://brainly.com/question/26146814

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