You heat 3.857 g of a mixture of Fe_3 O_4 and FeO to form 4.152 g Fe_2O_3. The m

Question

You heat 3.857 g of a mixture of Fe_3 O_4 and FeO to form 4.152 g Fe_2O_3. The mass percent of FeO Originally in the mixture was: 92.9% 38.7% 43.9% 56.1% none of these The following two reactions are important in the blast furnace production of iron metal from iron ore (Fe_2O_3): 2C(s) + O_2(g) rightarrow 2CO(g) Fe_2O_3(g) + 3CO(g) rightarrow 2Fe(e) + 3CO_2(g) Using these balanced reactions, how many moles of O_2 are required for the production of 5.15 kg of Fe? 30.7 moles 277 moles 92.2 moles 3.86 moles 69.2 moles

Explanation / Answer

The difference between the mass of the products and the massof the reactants has to be the mass of the oxygen taken from the air during the reaction (4.152g - 3.857g = 0.295g).

Moles of Oxygen= 0.295/32 =0.009219 .

Fe3O4 + FeO + 1/2 O2 ---->2Fe2O3

So moles of FeO =0.009219*2= 0.018438 moles mass of FeO in the sample =0.018438*72= 1.33

Mass % of FeO in the sample= 100*1.33/3.857=34.5%.    Hence close answer is 38.7%

2.

Moles of Fe in 5.15 kg= 5.15/56 kg moles =0.092 moles

Mole of CO required from 2nd reaction = 1.5 times Fe= 1.5*0.092= 0.138 kgmoles

From 1st reaction, mole of O2 required = 0.5 time moles of Oxygen =0.5*0.138 kg moles =0.069 kg moles = 0.069*1000= 69 kg moles

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