Respuesta :

BlueSky06

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.

The number of moles of fe is 27.4 g.

Calculation of no of moles:

Since

moles of Fe = 31.0 g  

So here the molar mass of Fe2O3 should be

= 2 x Fe (2 x 55.85) + 3 x O (3 x 16.00)

= 159.7 g/mole.

Now

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

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

And,

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

= 0.344 moles of Fe

Learn more about mole here: https://brainly.com/question/2200716

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