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This is an incomplete question for several reasons. Iron bromide can be FeBr2 or FeBr3 and this will influence the answer. Also, in order to precipitate silver bromide, one would add silver nitrate, but it isn't stated how much silver nitrate is used. Assuming FeBr2 is used, and there is excess AgNO3, then ....

2AgNO3 + FeBr2 ----> 2AgBr(s) + Fe(NO3)

moles FeBr2 = 2.96 g x 1 mol/216 g = 0.0137 moles

moles AgBr produced = 2 x 0.0137 = 0.0274 moles

mass AgBr produced = 0.0274 moles x 188 g/mole = 5.15 grams <---answer

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Q: What is the mass of silver bromide precipitated from 2.96 g of iron bromide?
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