The amount of [tex]Fe[/tex] found in the mixture of [tex]Fe_2O_3[/tex] and [tex]FeO[/tex] is 72.00 %. (Given)
So, in total mixture as 100%, 72.00 % of [tex]Fe[/tex] is present.
Molecular weight of [tex]Fe_2O_3[/tex] = [tex]55.845\times 2+15.99\times 3 = 159.66 g/mol[/tex]
Molecular weight of [tex]FeO[/tex] = [tex]55.845+15.99 = 71.835 g/mol[/tex]
Weight of mixture = [tex]71.835 +159.66 = 231.495 g[/tex]
Therefore, [tex]231.495 g[/tex] of mixture contains [tex]X g[/tex] of [tex]Fe[/tex].
So, [tex]X= 231.495\times \frac{72}{100} = 166.677 g[/tex]
That means [tex]231.495 g[/tex] of mixture contains [tex]166.677 g[/tex] of [tex]Fe[/tex]
So, [tex]0.0493 g[/tex] of [tex]Fe_2O_3[/tex] contains [tex]X g[/tex].
[tex]X= \frac{0.0493\times 166.677}{231.495} = 0.0355 g[/tex]
Hence, the mass of [tex]Fe_2O_3[/tex] in [tex]0.0493 g[/tex] of this mixture is [tex]0.0355 g[/tex].
