Explanation:
The given balanced reaction is as follows.
![2NH_{4}NO_{3} \rightarrow 2N_{2} + O_{2} + 4H_{2}O](https://tex.z-dn.net/?f=2NH_%7B4%7DNO_%7B3%7D%20%5Crightarrow%202N_%7B2%7D%20%2B%20O_%7B2%7D%20%2B%204H_%7B2%7DO)
It is given that mass of ammonium nitrate is 86.0 kg.
As 1 kg = 1000 g. So, 86.0 kg = 86000 g.
Hence, moles of
present will be as follows.
Moles of
= ![\frac{mass given}{\text{molar mass of NH_{4}NO_{3}}}](https://tex.z-dn.net/?f=%5Cfrac%7Bmass%20given%7D%7B%5Ctext%7Bmolar%20mass%20of%20NH_%7B4%7DNO_%7B3%7D%7D%7D)
= ![\frac{86000 g}{80.043 g/mol}](https://tex.z-dn.net/?f=%5Cfrac%7B86000%20g%7D%7B80.043%20g%2Fmol%7D)
= 1074.42 mol
Therefore, moles of
,
and
produced by 1074.42 mole of
will be as follows.
Moles of
= ![\frac{1}{2} \times 1074.42 mol](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes%201074.42%20mol)
= 537.21 mol
Moles of
= ![\frac{2}{2} \times 1074.42 mol](https://tex.z-dn.net/?f=%5Cfrac%7B2%7D%7B2%7D%20%5Ctimes%201074.42%20mol)
= 1074.42 mol
Moles of
= ![\frac{4}{2} \times 1074.42 mol](https://tex.z-dn.net/?f=%5Cfrac%7B4%7D%7B2%7D%20%5Ctimes%201074.42%20mol)
= 2148.84 mol
Therefore, total number of moles will be as follows.
537.21 mol + 1074.42 mol + 2148.84 mol
= 3760.47 mol
According to ideal gas equation, PV = nRT. Hence, calculate the volume as follows.
PV = nRT
1 atm \times V = 3760.47 mol \times 0.0821 L atm/mol K \times 580 K[/tex] (as
= 307 + 273 = 580 K)
V = 179066.06 L
Thus, we can conclude that total volume of the gas is 179066.06 L.