Answer to Question #75757 in General Chemistry for Nate

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)
How many grams of NH4NO3 are needed to produce 16.1 L of oxygen?

Answer 1

Find the amount of oxygen needed to produce under the conditions specified:
pV = νRT; T = 623 К; R = 0.082 (atm ×L)/(K×mol)
ν = pV/RT = (0.95 ×16.1)/(0.082 ×623) = 0.3 (mol)
According to the reaction equation 2 moles of NH4NO3 are needed to produce 1 mole of oxygen. Then 0.6 moles NH4NO3 are needed to produce 0.3 moles of oxygen.
Find the mass of 0.6 moles of NH4NO3 (M = 80 g/mole):
m = 0.6 × 80 = 48 (g)
Answer
48 grams of NH4NO3 are needed to produce 16.1 L of oxygen.