What is the mass of silver bromide agbr precipatated from 3.96 g of iron (iii) bromide febr3 ? febr3 + agno3 = agbr + fe(no3)3

First, write a balanced equation
FeBr3 + 3AgNO3 -> 3AgBr + Fe(NO3)3

Find the number of moles of FeBr3 we started with to find the number of moles of AfBr we end up with (we need to build a ratio)

3.96g FeBr3 / 295.55 g/mol * 3 mol AgBr / 1 mol FeBr3 * 187.77 g /mol = 7.55 g AgBr

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