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.
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
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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
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