How many moles of $N_{2}$ $N_2$ are formed by the decomposition of 2.5 moles of $N a N_{3}$ $NaN_3$ ?

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Answer 1 · 2016-02-16T19:40:56

$3.75 m o l N_{2}$ $3.75molN_2$

The decomposition reaction of sodium azide $N a N_{3}$ $NaN_3$ is:

$2 N a N_{3} (s) \to 2 N a (s) + 3 N_{2} (g)$ $2NaN_3(s)->2Na(s)+3N_2(g)$

Using dimensional analysis, we can find the number of mole of $N_{2}$ $N_2$ gas formed:

$?molN_2(g)=2.5cancel(mol NaN_3)xx(3molN_2)/(2cancel(molNaN_3))=3.75molN_2$

Answer 2 · 2016-02-16T19:44:03

$5.8 mols$

first multiply the mols with the mr $65*2.5=162.5$
then use this to get the mols of $N_2$
$mols=162.5/28$
$mols=5.8$

How many moles of N_2N2N_2 are formed by the decomposition of 2.5 moles of NaN_3NaN3NaN_3?