When iron and steam react at high temperatures, the following reaction takes place.3Fe(s) + 4H2O(g) → Fe3O4(s) + 4H2(g)How much iron must react with excess steam to form 300 g of Fe3O4(s) if the reaction yield is 69%?
Question
When iron and steam react at high temperatures, the following reaction takes place.3Fe(s) + 4H2O(g) → Fe3O4(s) + 4H2(g)How much iron must react with excess steam to form 300 g of Fe3O4(s) if the reaction yield is 69%?
Solution
To solve this problem, we need to follow these steps:
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First, we need to find the molar mass of Fe3O4. The molar mass of Fe is approximately 55.85 g/mol, and the molar mass of O is approximately 16.00 g/mol. Therefore, the molar mass of Fe3O4 is (355.85) + (416.00) = 231.55 g/mol.
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Next, we need to find the number of moles in 300 g of Fe3O4. We can do this by dividing the mass by the molar mass. So, 300 g / 231.55 g/mol = 1.295 mol.
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The balanced chemical equation tells us that 3 moles of Fe react to form 1 mole of Fe3O4. Therefore, to form 1.295 moles of Fe3O4, we need 3 * 1.295 = 3.885 moles of Fe.
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The molar mass of Fe is 55.85 g/mol, so 3.885 moles of Fe is 3.885 * 55.85 = 217.04 g.
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However, the problem states that the reaction yield is only 69%. This means that only 69% of the theoretical amount of Fe will actually react. Therefore, we need to divide the theoretical amount by the percentage yield to find the actual amount of Fe needed. So, 217.04 g / 0.69 = 314.61 g.
Therefore, approximately 315 g of Fe must react with excess steam to form 300 g of Fe3O4 if the reaction yield is 69%.
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