Answer
See the answer below.
Work Step by Step
Number of moles of oxygen:
$n=PV/RT$
$n=(735/760×1\ atm)×0.327\ L/((0.082)(19+273)\ K)$
$n=0.0132\ mol$
From stoichiometry, the number of moles of $KClO_3$ is:
$0.0132\ mol×2/3=0.0088\ mol$
This has a mass of:
$122.55\ g/mol×0.0088\ mol=1.08\ g$
Mass fraction:
$1.08\div1.56×100\%=69\%$
