Fe2O3 ==> 2Fe so you see there are 2 mols Fe in 2 mol Fe2O3.
To get 1100 kg Fe will require
1100 kg Fe x (1 mol Fe2O3/2 mols Fe) = about 550 kg Fe2O3. That is only 84% Fe2O3 in the ore itself; therefore, 550/0.84 = ? kg of the ore.
The rock in a particular iron ore deposit contains 84 % Fe2O3 by mass.
How many kilograms of the rock must be processed to obtain 1100 kg of iron?
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