Answer:
The mass of FeSO4*7H2O in the sample is 1.21 grams
Explanation:
<u>Step 1</u>: 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
<u>Step 2</u>: 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
<u>Step 3:</u> 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
<u>Step 4:</u> 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