Question

A powder contains FeSO4⋅7H2O (molar mass=278.01 g/mol), among other components. A 3.930 g sample of the powder was dissolved in HNO3 and heated to convert all iron to Fe3+. The addition of NH3 precipitated Fe2O3⋅xH2O, which was subsequently ignited to produce 0.348 g Fe2O3. What was the mass of FeSO4⋅7H2O in the 3.930 g sample?

Answer 1

Answer:

The mass of FeSO4*7H2O in the sample is 1.21 grams

Explanation:

Step 1: Calculate moles of Fe2O3

moles of Fe2O3 = mass of Fe2O3 / Molar mass of Fe2O3

moles of Fe2O3 = 0.348 grams / 159.69 g/mole = 0.00218 moles

Step 2: Calculate moles of Fe

4 Fe + 3O2 → 2Fe2O3

For 4 moles of Fe consumed there is 2 moles of Fe2O3 produced

This means it has a ratio 2:1

So 0.00218 moles of Fe2O3 produced , there is 2*0.00218 = 0.00436 moles of Fe consumed

Step 3: Calculate moles of FeSO4*7H2O

Fe + H2SO4 + 7H2O → FeSO4*7H20 + H2

For 1 mole of Fe consumed there is 1 mole of FeSO4*7H2O produced

This means for 0.00436 moles there is 0.00436 moles of Fe2SO4*H2O produced

Step 4: Calculate the mass of FeSO4*7H2O in the sample

mass of FeSO4*7H2O = 0.00436 moles * 278.01 g/mole = 1.212 g

The mass of FeSO4*7H2O in the sample is 1.21 grams