Oxygen, which is formed during the decomposition of 659.7 g of potassium permanganate containing 4.2%

Oxygen, which is formed during the decomposition of 659.7 g of potassium permanganate containing 4.2% of anoxic impurity, was used for the catalytic oxidation of ammonia. Calculate how much ammonia can react and how much (at standard) the nitrogen-containing product of this reaction will have.

2KMnO4 = K2MnO4 + MnO2 + O2
5O2 + 4NH3 = (k) 4NO + 6H2O
w (KMnO4 pure) = 100% -4.2% = 95.8%
m (KMnO4) = 659.7 g * 0.958 = 631.99 g
n (KMnO4) 631.99 g / 158 g / mol = 4 mol
n (O2) = 0.5n (KMnO4) = 2 mol
n (NH3) = (2 * 4) / 5 n (O2) = 1.6 mol
m (NH3) = 1.6 mol * 22.4 l / mol = 27.2 g
V (NO) = 1.6 mol * 22.4 L / mol = 35.84 L