Question

How many moles of N2 are formed by the decomposition of 1.60 mol of NaN3 ?

Answer 1

Answer:

The number of moles of nitrogen gas produced are, 2.25 moles.

Answer 2

Given:

Number of moles of $\[Na{N_3} = 1.60\]$ mol

To Find: Number of moles of $\[{N_2}\]$ formed by the decomposition of 1.60 mol of $\[Na{N_3}\]$

Solution:

The decomposition of $\[Na{N_3}\]$ can be given as:

$\[2Na{N_3}(s) \to 2Na(s) + 3{N_2}(g)\]$

From the above reaction, the decomposition of

2 mol of $\[Na{N_3}\]$ produces $=3$ mol of $\[{N_2}\]$

1 mol of $\[Na{N_3}\]$ produces $\[ = \frac{3}{2}\]$ mol of $\[{N_2}\]$

Therefore, the decomposition of

1.60 mol of $\[Na{N_3}\]$ produces $\[ = \frac{3}{2} \times 1.60\]$ mol of $\[{N_2}\]$

1.60 mol of $\[Na{N_3}\]$ produces $=2.4$ mol of $\[{N_2}\]$

Hence, the number of moles of $\[{N_2}\]$ formed by the decomposition of 1.60 mol of $\[Na{N_3}\]$ is $\[2.4\]$.