What is the volume of oxygen released during the thermal decomposition of 17.38 g

What is the volume of oxygen released during the thermal decomposition of 17.38 g of lead nitrate (2) Pb (NO3) 2 containing 5% impurities?

Given:
m tech. (Pb (NO3) 2) = 17.38 g
ω approx. = 5%

To find:
V (O2) -?

1) 2Pb (NO3) 2 => 2PbO + 4NO2 ↑ + O2 ↑;
2) ω (Pb (NO3) 2) = 100% – ω approx. = 100% – 5% = 95%;
3) m clean. (Pb (NO3) 2) = ω * m tech. / 100% = 95% * 17.38 / 100% = 16.511 g;
4) M (Pb (NO3) 2) = Mr (Pb (NO3) 2) = Ar (Pb) * N (Pb) + Ar (N) * N (N) + Ar (O) * N (O) = 207 * 1 + 14 * 2 + 16 * 6 = 331 g / mol;
5) n (Pb (NO3) 2) = m / M = 16.511 / 331 = 0.05 mol;
6) n (O2) = n (Pb (NO3) 2) / 2 = 0.05 / 2 = 0.025 mol;
7) V (O2) = n * Vm = 0.025 * 22.4 = 0.56 liters.

Answer: The O2 volume is 0.56 liters.