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Answer to Question #15734 in Other Chemistry for Porscha

Question #15734
When iron is exposed to oxygen (air), it rusts. How many grams of iron are needed in order to prepare 12.25g of Fe2O3?
Expert's answer
4Fe + 3O2 = 2Fe2O3
M(Fe2O3) = 2*56 + 3*16 = 112 + 48 = 160 g/mol
n(Fe2O3)
= m(Fe2O3) / M(Fe2O3) = 12.25 / 160 = 0.0766 mol
n(Fe) = 2*n(Fe2O3) = 0.1532
mol
m(Fe) = 0.1532 * 56 = 8.58 g

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