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What is the percent yield for iron if the reaction of 65.0 g of iron oxide produces 33.0 g of iron?
Wiki User
∙ 11y ago
Well, I ran the numbers for what you have in the question, and it didn't make sense, so I'm assuming you really mean 65g of iron yields 33g iron oxide. I'm also assuming you mean iron(III) oxide, which is iron's most common valence state when it reacts with oxygen, which means the product's formula would be Fe2O3.
With those things in mind, here's your balanced equation:
4Fe + 3O2 --> 2Fe2O3.
Step1: given 65g Fe, how many moles is that? Direct proportion between 65g and the unknown # of moles and iron's Atomic Mass of 56g/mol gives us 1.16 moles of iron.
Step 2: molar ratio of iron to iron(III) oxide is 4-2, which simplifies to 2-1, so you will have 0.58 moles of product.
Step 3: given 0.58 moles of product, set up another direct proportion with the unknown mass over given moles equal to iron(III) oxide's molar mass of 160g/mol to get a mass of 92.8g theoretical yield of product.
Step 4: your actual yield was 33g, so divide that by the stoichiometric yield of 92.8g, which is 0.355, which means about a 36% actual yield, which is the answer.
Wiki User
∙ 11y ago
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