The formula for rust can be represented by fe2o3. how many moles of fe are present in 27.4 g of the compound?

The solution would be like this for this specific problem:

Given:

moles of Fe = 31.0 g   So first, we get the molar mass of Fe2O3: The molar mass of Fe2O3 = 2 x Fe (2 x 55.85) + 3 x O (3 x 16.00) = 159.7 g/mole. 

27.4.0 g Fe2O3 x (1 mole Fe2O3 / 159.7 g Fe2O3) = 0.172moles of Fe2O3 

In the formula Fe2O3, there are 2 Fe. So, 1 mole of the compound Fe2O3 is composed of 2 moles of Fe. 

0.172 moles of Fe2O3 x (2 mole Fe / 1 mole Fe2O3) = 0.344 moles of Fe

Therefore, there are 0.344 moles of fe that are present in 27.4 g of the compound.

10.063g / 10.00 ml = 1.0063 g/ml which rounds off to 4 sig figs as: 1.006 g/ml

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