25.0 g TiO2 (1 mol TiO2/236.4 g)(2 mol Br2/3 mol TiO2)(159.8 g/1 mol Br2)
= 11.27 grams theoretical yield
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4.85/11.27 * 100
= 43% yield
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A mining company finds %TiO2 present in titanium ore, looking for the amount of bromine generated. 3TiO2 + 4BrF3 = 3TiF4 + 2Br2 + O2. If a 25.0 g sample of ore creates 4.85 g of Br2, what's the percent yield for TiO2 in the ore?