# Chapter 4 Stoichiometry: Quantitative Information about Chemical Reactions - Study Questions - Page 179c: 30

$8.80\%$

#### Work Step by Step

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: $2.491\ mmol\times26.982\ mg/mmol=67.2\ mg$ $67.2/764\times100\%=8.80\%$

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