Answer:
The mass fraction of ferric oxide in the original sample :[tex]\frac{723}{3110}[/tex]
Explanation:
Mass of the mixture = 3.110 g
Mass of [tex]Fe_2O_3=x[/tex]
Mass of [tex]Al_2O_3=y[/tex]
After heating the mixture it allowed to react with hydrogen gas in which all the ferric oxide reacted to form metallic iron and water vapors where as aluminum oxide did not react.
[tex]Fe_2O_3(s)+3H_2(g)\rightarrow 2Fe(s)+3H_2O(g)[/tex]
Mass of mixture left after all the ferric oxide has reacted = 2.387 g
Mass of mixture left after all the ferric oxide has reacted = y
[tex]x=3.110 g- y=3.110 g - 2.387 g = 0.723 g[/tex]
The mass fraction of ferric oxide in the original sample :
[tex]\frac{0.723 g}{3.110 g}=\frac{723}{3110}[/tex]