For the following reaction, 42.0 grams of iron are allowed to react with with 16

Question

For the following reaction, 42.0 grams of iron are allowed to react with with 16.0 grams of oxygen gas.

iron (s) + oxygen (g) iron(III) oxide (s)

What is the maximum amount of iron(III) oxide that can be formed? ___?___grams

What is the FORMULA for the limiting reagent?

What amount of the excess reagent remains after the reaction is complete? ___?___grams

Explanation / Answer

4Fe + 3O2 --> 2Fe2O3 Moles of Fe = .75 Moles of O2 = 16/32 = .5 4 moles need 3 moles .75 needs = .5625 O2 is limiting reagent so 3 moles give 2 moles Fe2O3 so .5 moles give 1/3 mole Maximum amount = 53.33 g if Fe2O3 Limiting reagent = O2 .083*56 = 4.67 gm

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