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OpenStudy (anonymous):

Fe2O3 + 2Al -> Al2O3 + 2Fe Calculate the mass of iron metal (in grams) that can be prepared from 150 grams of aluminum and 250 grams of iron(III) oxide. Help please!

OpenStudy (anonymous):

175

OpenStudy (anonymous):

can you explain a little bit please?

OpenStudy (anonymous):

Its a redox reaction: Basically the reaction is as follows: 2Al+Fe2O3→Al2O3+2Fe There are 150÷27.0=5.5556 moles of Aluminium,and 250÷(56×2+16×3)=1.5625 moles of iron (III) oxide. Which material is the limiting reagent? Very obviously, iron oxide is the limiting reagent, as there is less of that material. Thus, the mass of iron that can be prepared will be the amount of iron in 1.5625 mol of iron (III) oxide. I think you will be able to figure that out:)

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