Question

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)

Answer 1

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