Question

A chemist heats 100.0 g of es001-1.jpg in a crucible to drive off the water. If all the water is driven off, what is the mass of the remaining salt?

Answer 1

Answer:

FeSO₄*7H₂O(s) = FeSO₄(s) + 7H₂O(g)

M(FeSO₄*7H₂O)=278.0 g/mol

M(FeSO₄)=151.9 g/mol

m(FeSO₄*7H₂O)/M(FeSO₄*7H₂O)=m(FeSO₄)/M(FeSO₄)

m(FeSO₄)=M(FeSO₄)m(FeSO₄*7H₂O)/M(FeSO₄*7H₂O)

m(FeSO₄)=151.9*100.0/278.0=54.6 g

m(FeSO₄)=54.6 g

Answer 2

Answer:

FeSO4*7H2O(s) = FeSO4(s)+7H2O

M(FeSO4*7H2O) = 278g/mol

M (FeSO4)= 151.9g/mol

m(FeSO4*7H2O) / M(FeSO4*7H2O)= m(FeSO4)/M(FeSO4)

m(FeSO4)=M(FeSO4)m(FeSO4*7H2O)/M(FeSO4*7H2O)

m(FeSO4)= 151.9*100 / 278= 54.6g

m(FeSO4)= 54.6gram

The terms ferrous sulphate and iron(II) sulphate refer to a group of salts with the formula Fe SO4xH2O. These compounds are most frequently found as the heptahydrate (x = 7), while there are other known values for x. The hydrated form has industrial uses as well as medical uses for treating an iron deficit. The blue-green heptahydrate (hydrate with 7 molecules of water) is the most prevalent form of this substance, which has been known as copperas and green vitriol since antiquity. The identical aquo complex [Fe(H2O)6]2+, with octahedral molecular geometry and paramagnetic properties, results from the dissolution of all iron(II) sulphates in water.

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