Answer
$10.2\ g$
Work Step by Step
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:
$0.0760\ mol\times 134.11\ g/mol=10.2\ g$
