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For the following reaction, 0.481 moles of iron are mixed with 0.295 moles of oxygen gas. iron(s) + oxygen(g) → iron(II) oxid
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Answer #1

2 Fe (s) + O₂(g) 2 F e o (s) Fe = 0.481 mola o = 0.295 moles 2 molre need Imol O₂ 0.481 mol fe=1x0 481= 0.2405 onda [fe is li

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