Question

Na3N decomposes to form sodium and nitrogen gas at STP. If 13.7 L of nitrogen is produced

how many moles of Na3N was used? (22.4 L = 1 mole of any gas)
2Na3N --> 6Na + N2

Answer 1

Answer: 1.224 moles of
`Na_3N` were used.

Step-by-step 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 =
`(1mol)/(22.4L)* 13.7L=0.612mol`

For the given chemical reaction:

`2Na_3N\rightarrow 6Na+N_2`

By Stoichiometry of the reaction:

1 mole of nitrogen gas is produced by 2 mole of
`Na_3N`

So, 0.612 moles of nitrogen gas will be produced from =
`(2)/(1)* 0.612=1.224mol` of
`Na_3N`

Hence, 1.224 moles of
`Na_3N` were used.