Answer:
[tex]89.55~\%~of~Fe_2O_3~in~the~sample[/tex]
Explanation:
The first step in this reaction is the converstion from Kg of [tex]Fe[/tex] to grams of [tex]Fe_2O_3[/tex].
[tex]1.19x10^3~Kg~Fe~\frac{1000g~Fe}{1~Kg~Fe}~\frac{1~mol~Fe}{55.84~g~Fe}~\frac{1~mol~Fe_2O_3}{2~mol~Fe}~\frac{159.68~g~Fe_2O_3}{1~mol~Fe_2O_3}~\frac{1~Kg~Fe_2O_3}{1000~g~Fe_2O_3}[/tex]
[tex]X~=~1.7x10^3~Kg~Fe_2O_3[/tex]
Then we can calculate the percentage of [tex]Fe_2O_3[/tex] in the sample:
[tex]\%~Fe_2O_3~=~\frac{1.7x10^3~Kg~Fe_2O_3}{1.9x10^3~Kg~Sample}*100[/tex]
[tex]89.55~\%~of~Fe_2O_3~in~the~sample[/tex]