4.8 g O2 is mixed with 0.15 moles of iron. Given that iron (II) oxide will react with any available oxygen to form iron (III) oxide completely, the mass of iron (III) oxide formed would be :
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Answer:
4.8 g O2=4.832=0.15 Mol 0.15 mol O2 produces product =2×0.15=0.3 molFe2O3 mass of Fe2O3=159.692 gmol×0.3 mol = 47.9076 gm Regards.
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Answer:
2Feo + o ---> Fe2o3
Here oxigen is 4.8gm so mole of O2 will be 4.8÷32 that is 0.15 mol & given that Feo is 0.15 mol so Limiting Reagent is Feo.
From balanced chemical equation ---
2 mole of Feo forms 1mole of Fe2O3 .
therefore moles form by 0.15mol Feo =0.078
therefore mass of Fe2O3= 159×0.075=12gram
Explanation:
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