Question

When heated to 350 ∘C at 0.950 atm, ammonium nitrate decomposes to produce nitrogen, water, and oxygen gases: 2NH4NO3(s)→2N2(g)+4H2O(g)+O2(g)

a)How many liters of water vapor are produced when 26.4 g of NH4NO3 decomposes? Express your answer with the appropriate units.

b)How many grams of NH4NO3 are needed to produce 15.7 L of oxygen? Express your answer with the appropriate units.

Answer 1

a)

Molar mass of NH₄NO₃ = 80.043 g/mol

Moles of NH₄NO₃ = 26.4/80.043 = 0.330 mol

2NH₄NO₃(s)2N₂(g)+4H₂O(g)+O₂(g)

It is clear that 2 mol of NH₄NO₃ produces 4 mol of water

Therefore, 0.330 mol of NH₄NO₃ will produce = (4 x 0.330)/2 = 0.660 mol of water

From ideal gas law,

PV = nRT

where,

P = pressure = 0.950 atm

V = volume

n = number of moles of water vapor = 0.660 mol

R = gas constant = 0.0821 L.atm/mol.K

T = temperature = 350 ^∘C = 350 + 273 = 623 K

Now, substituting these values in the above equation to get,

0.950 x V = 0.660 x 0.0821 x 623

or, V = 35.5 L

Therefore, the volume of water vapor produced = 35.5 L

------

b)

For oxygen, given,

V = 15.7 L

P = 0.950 atm

T = 350 ^∘C = 350 + 273 = 623 K

R = 0.0821 L.atm/mol.K

Now, from ideal gas law,

PV = nRT

or, 0.95 x 15.7 = n x 0.0821 x 623

or, n = 0.292 mol

Now,

2NH₄NO₃(s)2N₂(g)+4H₂O(g)+O₂(g)

It is clear that 1 mol of O₂ is produced from 2 mol of NH₄NO₃

Therefore, 0.292 mol of O₂ is produced from = 2 x 0.292 = 0.584 mol of NH₄NO₃

Molar mass of NH₄NO₃ = 80.043 g/mol

Hence, the amount of NH₄NO₃ needed = 0.584 x 80.043 = 46.7 g