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 (NH_{4}NO_{3}) - ?

2NH_{4}NO_{3} → 2N_{2}↑ + O_{2}↑ + 4H_{2}O↑

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 (NH_{4}NO_{3}) = n (NH_{4}NO_{3}) * M (NH_{4}NO_{3}) = 0.724 mol * 80 g /mol = 57. 92 g ≈ 58 g

Condition:

t = 350 °C;

P = 0.950 atm

V (O2) = 19.5 L

m (NH

2NH

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 (NH

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