Answer: 1.224 moles of [tex]Na_3N[/tex] were used.
Explanation:
We are given:
Volume of nitrogen gas produced = 13.7 L
At STP conditions:
22.4 L of volume is occupied by 1 mole of a gas
Applying unitary method:
13.7 L of nitrogen gas will be occupied by = [tex]\frac{1mol}{22.4L}\times 13.7L=0.612mol[/tex]
For the given chemical reaction:
[tex]2Na_3N\rightarrow 6Na+N_2[/tex]
By Stoichiometry of the reaction:
1 mole of nitrogen gas is produced by 2 mole of [tex]Na_3N[/tex]
So, 0.612 moles of nitrogen gas will be produced from = [tex]\frac{2}{1}\times 0.612=1.224mol[/tex] of [tex]Na_3N[/tex]
Hence, 1.224 moles of [tex]Na_3N[/tex] were used.
