## Chemistry and Chemical Reactivity (9th Edition)

$10.2\ g$
Number of moles of Al: $2.05\ g\div 26.98\ g/mol=76.0\ mmol$ Number of moles of the base: $185\ mL\times 1.35\ M=250.\ mmol$ No aluminum will remain, since they both have the same stoichiometric coefficient. From stoichiometry: $0.0760\ mol$ of $KAl(OH)_4$ Mass produced: $0.0760\ mol\times 134.11\ g/mol=10.2\ g$