Final answer:
The mass of silver bromide precipitated from 2.96g of iron(III) bromide is 3.48g.
Step-by-step explanation:
To calculate the mass of silver bromide precipitated from iron(III) bromide, we need to use the information provided in the question. First, we need to find the molar mass of silver bromide by adding the atomic masses of silver and bromine. The molar mass of silver is 107.87 g/mol, and the molar mass of bromine is 79.9 g/mol. So, the molar mass of silver bromide is 107.87 + (3 * 79.9) = 347.57 g/mol.
Next, we need to convert the given mass of iron(III) bromide to moles. The molar mass of iron(III) bromide is 295.63 g/mol. We can calculate the number of moles of iron(III) bromide by dividing the given mass by its molar mass: 2.96 g / 295.63 g/mol = 0.01 mol.
Since the stoichiometry of the reaction is 1:1 between iron(III) bromide and silver bromide, the mass of silver bromide precipitated is also 0.01 mol. To find the mass, we multiply the number of moles by the molar mass of silver bromide: 0.01 mol * 347.57 g/mol = 3.48 g. Therefore, the mass of silver bromide precipitated from 2.96 g of iron(III) bromide is 3.48 g.