Respuesta :
Explanation:
The given balanced reaction is as follows.
[tex]2NH_{4}NO_{3} \rightarrow 2N_{2} + O_{2} + 4H_{2}O[/tex]
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 [tex]NH_{4}NO_{3}[/tex] present will be as follows.
Moles of [tex]NH_{4}NO_{3}[/tex] = [tex]\frac{mass given}{\text{molar mass of NH_{4}NO_{3}}}[/tex]
= [tex]\frac{86000 g}{80.043 g/mol}[/tex]
= 1074.42 mol
Therefore, moles of [tex]N_{2}[/tex], [tex]O_{2}[/tex] and [tex]H_{2}O[/tex] produced by 1074.42 mole of [tex]NH_{4}NO_{3}[/tex] will be as follows.
Moles of [tex]O_{2}[/tex] = [tex]\frac{1}{2} \times 1074.42 mol[/tex]
= 537.21 mol
Moles of [tex]N_{2}[/tex] = [tex]\frac{2}{2} \times 1074.42 mol[/tex]
= 1074.42 mol
Moles of [tex]H_{2}O[/tex] = [tex]\frac{4}{2} \times 1074.42 mol[/tex]
= 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 [tex]307^{o}C[/tex] = 307 + 273 = 580 K)
V = 179066.06 L
Thus, we can conclude that total volume of the gas is 179066.06 L.
