A powder contains FeSO4 · 7 H2O (molar mass = 278.01 g/mol), among other components. A 3.055 g sample of the powder was dissolved in HNO, and heated to convert all iron to Fe3+. The addition of NH, precipitated Fe, Oz · xH, O, which was subsequently ignited to produce 0.294 g Fe2O3. What was the mass of FeSO4 · 7H2O in the 3.055 g sample?

The mass of FeSO4*7H2O is 1.023 grams

Explanation:

Step 1: Data given

Molar mass of FeSO4 * 7H2O = 278.01 g/mol

Mass of Fe2O3 = 0.294 grams

Mass of the sample = 3.055 grams

Step 2: The balanced equation

4 Fe + 3O2 → 2Fe2O3

Step 3: Calculate moles Fe2O3

Moles Fe2O3 = mass Fe2O3 / molar mass Fe2O3

Moles Fe2O3 = 0.294 grams / 159.59 g/mol

Moles Fe2O3 = 0.00184 moles

Step 3: Calculate moles Fe

In 2 moles Fe2O3 we have 4 moles Fe

For 0.00184 moles we'll have 2*0.00184 = 0.00368 moles

Step 4: The balanced equation

Fe + H2SO4 + 7H2O → FeSO4*7H20 + H2

Step 5: Calculate moles FeSO4 * 7H2O

For 1 mol Fe we have 1 mol FeSO4*7H2O

For 0.00368 moles Fe we have 0.00368 moles FeSO4*7H2O

Step 6: Calculate the mass of FeSO4*7H2O

Mass FeSO4*7H2O = moles * molar mass

Mass FeSO4*7H2O = 0.00368 moles * 278.01 g/mol

Mass FeSO4*7H2O = 1.023 grams

The mass of FeSO4*7H2O is 1.023 grams