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 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