A powder contains FeSO4⋅7H2O (molar mass=278.01 g/mol), among other components. A 2.810 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 0.443 g Fe2O3.What was the mass of FeSO4⋅7H2O in the 2.810 g sample?

the mass of FeSO4.7H2O  in  the 2.810 g sample was 1.5402 g

Explanation:

From the given information:

Two  moles of FeSO4.7H2O = one mole of  Fe2O3

Let recall that:

number of moles of  Fe2O3 = mass of  Fe2O3 / molar mass of  Fe2O3

Given that :

mass of Fe2O3 = 0.443 g

number of moles of  Fe2O3 = 0.443 g/ 159.69 g/mol

number of moles of  Fe2O3 = 0.00277 mol

Thus;

number of moles of FeSO4.7H2O = 2 ×  Fe2O3

number of moles of FeSO4.7H2O = 2 × 0.00277 mol

number of moles of FeSO4.7H2O = 0.00554 mol

However from the usual stoichiometry formula; the mass of a substance = number of moles  × molar mass

Now; the mass of FeSO4.7H2O = number of moles × molar mass

the mass of FeSO4.7H2O = 0.00554 mol × 278.01 g/mol

the mass of FeSO4.7H2O = 1.5402 g

Therefore; the mass of FeSO4.7H2O  in  the 2.810 g sample was 1.5402 g

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and vice versa holds true for equatorial positions

so the correct answer with labelled diagram will be