Question

How many grams of NH4NO3 are needed to produce 19.5 L of oxygen?
Express your answer with the appropriate units.

m=??

Answer 1

Answer: 58 g
Condition:
t = 350 °C;
P = 0.950 atm
V (O2) = 19.5 L
m (NH₄NO₃) - ?
2NH₄NO₃ → 2N₂↑ + O₂↑ + 4H₂O↑
Solution:
PV = nRT; n = PV / RT
T (O2) = 350 °C + 273 = 623 K
n (O2) = 0.950 atm * 19.500 L / (0.0821 L * atm /mol * K) * 623 K = 18.525 / 51.148 = 0.362 mol
n (NH4NO3) = n (O2) * 2 = 0.362 mol * 2 = 0.724 mol
m (NH₄NO₃) = n (NH₄NO₃) * M (NH₄NO₃) = 0.724 mol * 80 g /mol = 57. 92 g ≈ 58 g