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I got the same answer: 12 mols ClO2 (looking at the ratio of 6 ClO2: 6 ClO2F) and 2 mols of Br2 (again, looking at the ratio 2BrF3: 1Br2). Since we know that BrF3 is the limiting reactant and that there are 4 mol of BrF3 needed for 12 mols ClO2, we know that we have 1 mol excess of BrF3 after 2 moles of Br2 are produced.
That is the correct answer. I simply recognized that with 12 moles of ClO2, I needed 4 moles of BrF3 and this would produce double the typical amount of each product (as double the ClO2 and BrF3 was used) so I got 12 mols ClO2F as a product as well as 2 mols Br. Since I only used 4 mole of BrF3, there should be one excess mole remaining.
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