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Answer 1

Answer #1

Molar mass of Fe2O3= 2 * 55.84 + 3*16.0 =111.68 + 48.0 = 159.68 gm Fe2O3

111.68 gm of Fe is present in 159.68 gm of Fe2O3

how much Fe present in 21.5 gm of Fe2O3

amount of Fe present = (21.5*111.68)/159.68 = 15 gm of Fe

Moles of Fe present = (15/55.84) = 0.268 mol

Answer 2

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