## Chemistry and Chemical Reactivity (9th Edition)

$22.74\ g\ Br_2$ $25.31\ g\ Al_2Br_6$
Number of moles of aluminum: $2.56\ g\div26.982\ g/mol=0.0949\ mol$ From stoichiometry: $0.0949\ mol\times3/2\times159.808\ g/mol\ Br_2=22.74\ g\ Br_2$ $0.0949\ mol\times1/2\times 533.387\ g/mol\ Al_2Br_6=25.31\ g\ Al_2Br_6$