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A powder contains FeSO4 7 H,O (molar mass = 278.01 g/mol), among other components. A 3.995 g sample of the powder was dissolv

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Answer #1

mass of FeSO4.7H2O = 0.731 g

Explanation

mass Fe2O3 = 0.210 g

moles Fe2O3 = (mass Fe2O3) / (molar mass Fe2O3)

moles Fe2O3 = (0.210 g) / (159.69 g/mol)

moles Fe2O3 = 1.315 x 10-3 mol

moles FeSO4.7H2O = 2 * (moles Fe2O3)

moles FeSO4.7H2O = 2 * (1.315 x 10-3 mol)

moles FeSO4.7H2O = 2.63 x 10-3 mol

mass FeSO4.7H2O = (moles FeSO4.7H2O) * (molar mass FeSO4.7H2O)

mass FeSO4.7H2O = (2.63 x 10-3 mol) * (278.01 g/mol)

mass FeSO4.7H2O = 0.731 g

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