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Answer to Question #12739 in Other Chemistry for Molly

Question #12739
What is the mass of iron(III) bromide (295.55 g/mol) that yields 0.188 g of silver bromide (187.77 g/mol) precipitate?

__FeBr3(s) + __AgNO3(aq) → __AgBr(s) + __Fe(NO3)3(aq)
Expert's answer
m 0.188
FeBr3(s) + 3AgNO3(aq) = 3AgBr(s) + Fe(NO3)3(aq)
295.55 3*187.77

m = 295.55*0.188/(3*187.77) = 0.986 g

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