Answer:
The answer to your question is: 53.46 % pure
Explanation:
data
Fe = 1.92 ✕ 103 kg  produced   = 1920 kg
Fe2O3 = 5.13 ✕ 103 kg   sample  = 5130 kg
MW Fe2O3 = (56x2)+(16x3) = 160 kg
% of purity = ?
          Fe2O3 + 3 CO → 2 Fe + 3 CO2
Convert mass to moles
Fe
   56 kg  --------------------- 1 mol
 1920 kg ---------------------  x moles    x = 34.28 moles
From the reaction
  1 mol of Fe2O3 ---------------------  2 moles of Fe
   x moles of Fe2O3 --------------- 34.28 moles
 x = 34.28/2 = 17.14 moles of Fe2O3
    160 kg of FE2O3 ----------------  1 mol
       x kg of Fe2O3---------------  17.14 moles
x = 17.14 x 160/1 = 2742,4 kg of Fe2O3 Â It's supposed to be the amount of Fe if it was 100% pure.
               2742.4 kg of Fe2O3 ----------------  100%
               5130 kg of Fe2O3 -------------------  x
x = (2742.4x100)/5130 = 53.46 pure
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