Answer:
[tex]6molNa\\\\9molN_2[/tex]
Explanation:
Hello there!
In this case, given the chemical reaction:
[tex]2NaN_3\rightarrow 2Na+3N2[/tex]
It is possible to evidence the 2:2 mole ratio of the reactant to Na and the 2:3 mole ratio of the reactant to the N2; therefore, the following setups allows us to compute the moles of products:
[tex]6molNaN_3*\frac{2molNa}{2molNaN_3}=6molNa\\\\6molNaN_3*\frac{3molN_2}{2molNaN_3}=9molN_2[/tex]
Best regards!
