Question

The rapid decomposition of sodium azide, NaN3, to its elements is one of the reactions used to inflate airbags: 2 NaN3 (s)  2 Na (s) + 3 N2 (g) How many grams of N2 are produced from 6.00 g of NaN

Answer 1

Answer: The amount of nitrogen gas produced is 3.864 grams

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$     .....(1)

Given mass of $NaN_3$ = 6.00 g

Molar mass of $NaN_3$ = 65 g/mol

Putting values in equation 1, we get:

$\text{Moles of }NaN_3=\frac{6.00g}{65g/mol}=0.092mol$

For the given chemical reaction:

$2NaN_3(s)\rightarrow 2Na(s)+3N_2(g)$

By Stoichiometry of the reaction:

2 moles of $NaN_3$ produces 3 moles of nitrogen gas

So, 0.092 moles of $NaN_3$ will produce = $\frac{3}{2}\times 0.092=0.138mol$ of nitrogen gas

Now, calculating the mass of nitrogen gas by using equation 1, we get:

Molar mass of nitrogen gas = 28 g/mol

Moles of nitrogen gas = 0.138 moles

Putting values in equation 1, we get:

$0.138mol=\frac{\text{Mass of nitrogen gas}}{28g/mol}\\\\\text{Mass of nitrogen gas}=(0.138mol\times 28g/mol)=3.864g$

Hence, the amount of nitrogen gas produced is 3.864 grams