Answer
See answer below.
Work Step by Step
Number of moles of the salt:
$0.35\ M\times 15.0\ L=5.25\ mol$
From stoichiometry:
$5.25\ mol$ of NaOH,
$2.63\ mol$ of $Cl_2$
Masses formed:
$5.25\ mol\times40.0\ g/mol=210\ g$ of NaOH,
$2.63\ mol\times70.91\ g/mol=186\ g$ of $Cl_2$
