## Chemistry and Chemical Reactivity (9th Edition)

$8.80\%$
Number of moles of the oxide: $127\ mg\div 101.961\ mg/mmol=1.255\ mmol$ From stoichiometry: $1.255\ mmol\times 2_{Al(OH)_3}/1_{Al_2O_3}\times 1_{Al}/1_{Al(OH)_3}=2.491\ mmol$ of Al Mass fraction: $2.491\ mmol\times26.982\ mg/mmol=67.2\ mg$ $67.2/764\times100\%=8.80\%$