Question

What is the mass of silver bromide AgBr precipatated from 3.96 g of iron (III) bromide FeBr3 ? FeBr3 + AgNO3 = AgBr + Fe(NO3)3

Answer 1

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