For the oxidation of ammonia, such a volume of oxygen was required, which is formed during the decomposition

For the oxidation of ammonia, such a volume of oxygen was required, which is formed during the decomposition of 245.6 g of potassium permanganate containing 3.5% of impurities. Determine the mass of ammonia reacted and the volume of nitrogen formed.

2KMnO4 = K2MnO4 + MnO2 + O2

m (KMnO4) = w (KMnO4) * m (KMnO4 + impurities) = 0.965 * 245.6 = 237 g;

n (KMnO4) = m (KMnO4) / M (KMnO4) = 237/158 = 1.5 mol;

n (O2) = 0.5n (KMnO4) = 0.75 mol

2NH3 + 3O2 = N2 + 3H2O

n (NH3) = 2/3 n (O2) = 0.5 mol;

m (NH3) = n (NH3) * M (NH3) = 0.5 * 17 = 8.5 g;

n (N2) = 1/3 n (O2) = 0.25 mol;

V (N2) = 0.25 * 22.4 = 5.6 liters.

Answer: m (NH3) = 8.5 g; V (N2) = 5.6 liters.