Answer
$m(NaN_{3})=61.98g$
Work Step by Step
$V=25l=25dm^{3}$
$p=1.4atm=1.4\times 101.3kPa = 141.82 kPa$
$T=25^{\circ}C=(25+273.15)K=298.15K$
---$pV=n(N_{2})RT$
$n(N_{2})=\frac{pV}{RT}=\frac{141.82kPa\times 25dm^{3}}{8.314\frac{J}{K\times mol}\times 298.15K}=1.4303mol$
Let's consider the following reaction:
$2NaN_{3} \rightarrow 2Na + 3N_{2}$
In order to obtain $3$ moles of $N_{2}$, $2$ moles of $NaN_{3}$ are required (stoichiometric coefficients). Hence, we have the following proportion:
$2mol:3mol=n(NaN_{3}):n(N_{2})$
$\implies n(NaN_{3})=\frac{2}{3}n(N_{2})=0.9535mol$
$\implies m(NaN_{3})=M(NaN_{3})\times n(NaN_{3})=65\frac{g}{mol}\times 0.9535 mol=61.98g$
