Answer:
2.29 g of N2
Explanation:
We have to start with the chemical reaction:
[tex]NaN_3~->~Na~+~N_2[/tex]
The next step is to balance the reaction:
[tex]2NaN_3~->~2Na~+~3N_2[/tex]
We can continue with the mol calculation using the molar mass of
[tex]NaN_3[/tex] (65 g/mol), so:
[tex]3.55~g~NaN_3\frac{1~mol~NaN_3}{65~g~NaN_3}=0.054~mol~NaN_3[/tex]
Now, with the molar ratio between [tex]NaN_3[/tex] and [tex]N_2[/tex] we can calculate the moles of [tex]N_2[/tex] (2:3), so:
[tex]0.054~mol~NaN_3\frac{3~mol~N_2}{2~mol~NaN_3}=0.0819~mol~N_2[/tex]
With the molar mass of [tex]N_2[/tex] we can calculate the grams:
[tex]0.0819~mol~N_2=\frac{1~mol~N_2}{28~g~N_2}=2.29~g~N_2[/tex]
I hope it helps!