From an initial mix of: 2 mol Cl2, 1 mol Br2, 0.5 mol I2, 0.5 mol Cl−, 1 mol Br−, & 2 mol I−; the total quantity of I2 present after all possible reactions occur is
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Expert's answer
2015-12-02T13:42:19-0500
0.5 mol I2 and 0.5 mol Cl- will not react with anything; 2 mol I- will react with one mol of Cl2 producing one mol of I2: 2I- + Cl2 = I2 + 2Cl- 1 mol Br- will react with 0.5 mol of Cl2 producing 0.5 mol of Br2: Br- + 0.5Cl2 = 0.5 Br2 + Cl- Final mixture: 0.5 mol Cl2, 1.5 mol Br2, 1.5 mol I2, 3.5 mol Cl- Quantity of I2: 1.5 mol
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