Question & Answer: Iron reacts with 48.0 g of oxygen to make 159.6 g of rust (Fe_2O_3): how much iron reacted?…..

Iron reacts with 48.0 g of oxygen to make 159.6 g of rust (Fe_2O_3): how much iron reacted? 111.6 g 48.0 g 207.6 g 55.8 g

Expert Answer

Answer

4 Fe + 3O₂ ————> 2 Fe₂O₃

Molecular weight of Fe₂O₃ = 159.6 g

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Question & Answer: Iron reacts with 48.0 g of oxygen to make 159.6 g of rust (Fe_2O_3): how much iron reacted?…..

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Iron and oxygen react in 1:1.5 ratio to form the iron oxide

3 moles of oxygen required 2 moles of iron

Therefore iron required is 2 moles = 2 x 55.84 = 111.6 g

Therefore the answer is first option.

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