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
