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Question

For the following reaction, 44.3 grams of iron are allowed to react with 24.1 grams of oxygen gas .

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

What is the maximum mass of iron(III) oxide that can be formed?  grams

What is the FORMULA for the limiting reagent?

What mass of the excess reagent remains after the reaction is complete?  grams

Answer 1

Balance the reaction for mole ratios: 4 Fe + 3 O₂ --> 2 Fe₂O₃.

For iron: (44.3g Fe) * (1 mole Fe /55.845g Fe) * (2 mole Fe₂O₃ /4 mole Fe) = 0.396₆ mole Fe₂O₃.

For oxygen: (24.1g O₂) * (1 mole O₂ /32.00g O₂) * (2 mole Fe₂O₃ /3 mole O₂) = 0.502₁ mole Fe₂O₃.

Take lower value of moles of product: (0.396₆ mole Fe₂O₃) * (159.69g Fe₂O₃ /1 mole Fe₂O₃) = 63.3g Fe₂O₃.

Formula of limiting reagent: Fe.

Amount of excess reagent: (Δmole = 0.105₅ mole Fe₂O₃) * (3 mole O₂ /2 mole Fe₂O₃) * (32.00g O₂ /1 mole O₂) = 5.1g excess O₂.

Answer 2

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