Asked by Aria
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?
How many kilograms of the rock must be processed to obtain 1100 kg of iron?
Answers
Answered by
DrBob222
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.
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.
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