How many grams of NH4NO3 are needed to produce 19.5 L of oxygen?
Express your answer with the appropriate units.
m=??
1
Expert's answer
2015-11-03T13:51:38-0500
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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