Answer
See the answer below.
Work Step by Step
Number of moles of nitrogen, from ideal gas law:
$P.V=n.R.T$
$n=1.3\ atm×75.0\ L/0.082057\ L.atm/mol.K/(25+273)\ K$
$n=3.99\ mol$
From stoichiometry: $2.66\ mol$ of $NaN_3$
Mass of sodium azide:
$2.66\ mol×65.01\ g/mol=173\ g$