Question

Ammonium nitrate, a common fertilizer, is used as an explosive in fireworks and by terrorists. It was the material used in the tragic explosion at the Oklahoma City federal building in 1995. How many liters of gas at 307°C and 1.00 atm are formed by the explosive decomposition of 86.0 kg of ammonium nitrate to nitrogen, oxygen, and water vapor?

Answer 1

Step-by-step explanation:

The given balanced reaction is as follows.

`2NH_(4)NO_(3) \rightarrow 2N_(2) + O_(2) + 4H_(2)O`

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
`NH_(4)NO_(3)` present will be as follows.

Moles of
`NH_(4)NO_(3)` =
`\frac{mass given}{\text{molar mass of NH_(4)NO_(3)}}`

=
`(86000 g)/(80.043 g/mol)`

= 1074.42 mol

Therefore, moles of
`N_(2)`,
`O_(2)` and
`H_(2)O` produced by 1074.42 mole of
`NH_(4)NO_(3)` will be as follows.

Moles of
`O_(2)` =
`(1)/(2) * 1074.42 mol`

= 537.21 mol

Moles of
`N_(2)` =
`(2)/(2) * 1074.42 mol`

= 1074.42 mol

Moles of
`H_(2)O` =
`(4)/(2) * 1074.42 mol`

= 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^(o)C` = 307 + 273 = 580 K)

V = 179066.06 L

Thus, we can conclude that total volume of the gas is 179066.06 L.