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Answer to Question #56050 in General Chemistry for sue

Question #56050
How many grams of NH4NO3 are needed to produce 19.5 L of oxygen?
Express your answer with the appropriate units.

m=??
Expert's answer
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

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