In a 1.143e+2 g sample of iron(iii) oxide, how many grams of iron are present? your answer should have the appropriate number of significant figures (use two decimal places in atomic masses).

Mass of iron(III)oxide, Fe2O3 sample = 1.143*10^2 g

Molar mass of Fe2O3 = 159.69 g/mole

i.e. 1 mole of Fe2O3 weighs 159.69 g

Based on stoichiometry:

1 mole of Fe2O3 has 2 moles of Fe

Atomic mass of Fe = 55.85 g/mole

Therefore we can write:

159.69 g of Fe2O3 is composed of 2*55.85 g of Fe

Thus, amount of Fe present in 1.143*10^2 g sample of Fe2O3 is

= 1.143*10^2 * 2* 55.85/159.69

= 7.995*10^-1 g