What mass of Br2 is needed to react with 27 g of aluminum? 2Al + 3Br2 → 2AlBr3

Answer 1

2Al + 3Br₂ ==> 2AlBr₃

moles Al present = 27 g x 1 mole Al/27 g = 1 mole Al

mass Br₂ needed = 1 mole Al x 3 mole Br₂/mole Al x 160 g/mole Br₂ = 480 g Br₂ needed