If 130g of sodium azide is added to an air bag, what mass of gas can be produced by this reaction?

Steps to Solve

1. Identify the Chemical Reaction The decomposition reaction of sodium azide ($\text{NaN}_3$) can be represented as:
   $$ 2 , \text{NaN}_3 , (s) \rightarrow 3 , \text{N}_2 , (g) + \text{Na} , (s) $$

This tells us that 2 moles of sodium azide produce 3 moles of nitrogen gas.

2. Calculate Moles of Sodium Azide We need to determine the number of moles in 130 grams of sodium azide.
   The molar mass of sodium azide ($\text{NaN}_3$) is calculated as follows:

• Sodium (Na): 22.99 g/mol
• Nitrogen (N): 14.01 g/mol
  $$ \text{Molar mass of NaN}_3 = 22.99 , \text{g/mol} + 3 \times 14.01 , \text{g/mol} = 65.00 , \text{g/mol} $$

Now, calculate moles: $$ \text{Moles of NaN}_3 = \frac{130 , \text{g}}{65.00 , \text{g/mol}} = 2.00 , \text{mol} $$

3. Use Stoichiometry to Find Moles of Gas Produced Using the stoichiometry from the reaction, we find that 2 moles of sodium azide yield 3 moles of nitrogen gas. Thus for 2 moles of sodium azide, the moles of nitrogen gas produced will be: $$ \text{Moles of } \text{N}_2 = 2.00 , \text{mol} \times \frac{3 , \text{mol} , \text{N}_2}{2 , \text{mol} , \text{NaN}_3} = 3.00 , \text{mol} $$

4. Calculate the Mass of Gas Produced The mass of nitrogen gas can be calculated using its molar mass (28.02 g/mol):
   $$ \text{Mass of } \text{N}_2 = \text{Moles of } \text{N}_2 \times \text{Molar mass of } \text{N}_2 $$
   $$ \text{Mass of } \text{N}_2 = 3.00 , \text{mol} \times 28.02 , \text{g/mol} = 84.06 , \text{g} $$