In older airbags a crash triggers the decomposition of NaN3 to form elemental sodium and nitrogen gas, which fills the airbag. How many moles of N2 gas are produced when 154.7 g NaN3 decomposes?

Question

Whole Question: The deployment of airbags during a car accident results from a chemical reaction that, until the 1990s, involved sodium azide, NaN₃. Sodium azide has since been replaced by less toxic substances that react in a similar way. In older airbags a crash triggers the decomposition of NaN₃ to form elemental sodium and nitrogen gas, which fills the airbag. How many moles of N₂ gas are produced when 154.7g NaN₃ decomposes?



If possible, can I get a step by step solution? Thank you!

Answer 1

First you need to write out the chemical equation for NaN3 decomposition.

The general equation is:

NaN3 -> Na + N₂

Then you need to balance the equation.

In this case the balanced equation ends up being:

2 NaN₃ -> 2Na + 3N₂

Then we need to figure out the molecular weight of NaN₃

Which is: 22.99+ (3*14.01)= 65.02 g/mol

to find out how many moles of N₂

you need to divide the number off grams you start out with by molecular weight. 154.7/65.02 = 2.38 moles of NaN₃

To determine overall moles of N₂ gas you need to use the molar ratio we found in the equation.

2.38 moles NaN₃* (3 moles N₂/ 2 moles of NaN₃) = 3.57 moles of N₂

** make sure during all of the equations you are keeping track measurements