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3 years ago

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Chemistry

1 answer:

Licemer1 [7]3 years ago

3 0

Answer:

The answer to your question is:

Vol of NO2 = 11.19 L

Vol of O2 = 2.8 L

Explanation:

Data

N2O5 = 56 g

STP T = 0°C = 273°K

P = 1 atm

MW N2O5 = 216 g

Gases law = PV = nRT

Process

216 g of N2O5 ---------------- 1 mol

54 g ----------------- x

x = (54 x 1) / 216

x = 0.25 mol of N2O5

2 mol of N2O5 ----------------- 4 mol of NO2

0.25 mol ------------------ x

x = (0.25 x 4) / 2 = 0.5 mol of NO2

V = nRT/P

V = (0.5)(0.082)(273) / 1 = 11.19 L

2 mol of N2O5 ----------------- 1 O2

0.25 N2O5 ---------------------- x

x = (0.25 x 1) / 2 = 0.125 mol

Vol = (0.125)((0.082)(273) / 1 = 2.8 L

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