Question

A mixture of fe2o3 and feo was found to contain 72.00% fe by mass. what is the mass of fe2o3 in 0.0493 g of this mixture?

Answer 1

The amount of
`Fe` found in the mixture of
`Fe_2O_3` and
`FeO` is 72.00 %. (Given)

So, in total mixture as 100%, 72.00 % of
`Fe` is present.

Molecular weight of
`Fe_2O_3` =
`55.845* 2+15.99* 3 = 159.66 g/mol`

Molecular weight of
`FeO` =
`55.845+15.99 = 71.835 g/mol`

Weight of mixture =
`71.835 +159.66 = 231.495 g`

Therefore,
`231.495 g` of mixture contains
`X g` of
`Fe`.

So,
`X= 231.495* (72)/(100) = 166.677 g`

That means
`231.495 g` of mixture contains
`166.677 g` of
`Fe`

So,
`0.0493 g` of
`Fe_2O_3` contains
`X g`.

`X= (0.0493* 166.677)/(231.495) = 0.0355 g`

Hence, the mass of
`Fe_2O_3` in
`0.0493 g` of this mixture is
`0.0355 g`.