# How many moles of N_2 are produced by the decomposition of 1.25 mol of sodium azide in the reaction 2NaN_3(s) -> 2Na(s) + 3N_2(g)?

$\text{1.25 mol NaN"_3}$ will produce $\text{1.88 mol N"_2}$.
"2NaN"_3("s")$\rightarrow$$\text{2Na("s")+3N"_2("g")}$
Multiply the given moles of sodium azide by the mole ratio between $\text{NaN"_3}$ and $\text{N"_2}$ from the balanced equation, with $\text{N"_2}$ in the numerator. ${\text{2 mol N}}_{2} :$$\text{3 mol N"_2}$
$1.25 \cancel{\text{mol NaN"_3xx(3"mol N"_2)/(2cancel"mol NaN"_3)="1.88 mol N"_2}}$ rounded to three significant figures