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Chemistry 10 Online
OpenStudy (anonymous):

How much iron is present in 2 . 99 g of iron(III) oxide

OpenStudy (kira_yamato):

Proportion by mass Fe = 55.8(2)/(55.8(2) + 3(16.0) = 279/319 Mass of Fe = 2.99*279/319 ≈ 2.62 g

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