How many moles are in 79.85g fe2o3?

in 79.85 g of Fe2O3 there will be 0.5 moles of Fe2O3

Explanation:

To find the number of moles we must calculate the molecular weight of iron oxide, for that we use the periodic table:

mFe = 55.845 g/mol

mO = 15.999 g/mol

mFe2O3 = (2x55.845) + (3x15.999) = 159.687 g/mol

1 mole of iron oxide weighs 159.687 g, therefore, to calculate the number of moles present in 79.85 g it will be equal to:

nFe2O3 = 79.85/159.687 = 0.5 moles