Answer to Question #99055 in General Chemistry for Tyyana

Question

What is the mass of silver bromide precipitated frromm 2.96g of iron bromide

Answer 1

The reaction can be shown as follows:

FeBr₃ + 3Ag⁺ = 3AgBr + Fe³⁺

From here, the mass of precipitated silver bromide equals:

m(AgBr) = m(FeBr3) × 3 × Mr(AgBr) / Mr (FeBr₃) = 2.96 g × 3 × 188 g/mol / 296 g/mol = 5.64 g

Answer: 5.64 g of silver bromide