Mixed equal volumes of nitric oxide (II) and oxygen. After completion of the reaction, the gas mixture was passed through

Mixed equal volumes of nitric oxide (II) and oxygen. After completion of the reaction, the gas mixture was passed through an alkali solution. How and how much the volume of the gas mixture will change: a) after mixing nitrogen oxide (II) and oxygen; b) after passing the resulting mixture through an alkali solution?

1. Let’s compose the equation of the first reaction:
2NO + О2 = 2NO2 (1)
2. According to the law of volumetric ratios, the volumes of gases in reaction (1) relate to each other as coefficients:
V (NO): V (O): V (NO2) = 2: 1: 2.
Let the initial mixture contain 1 volume of NO and O2.
0.5 volume of O2 reacts with 1 volume of NO;
1 volume of NO2 is formed,
0.5 volume of O2 remains.
3. Thus, after mixing, the volume of the gas mixture decreased to 1.5 volumes.
4. Let us write the equation for the reaction of NO2 with alkali in the presence of oxygen:
4NO2 + 4KON + O2 = 4KNO3 + 2H2O (2)
5. Let us find the volume of oxygen reacted with 1 volume of NO2 according to the equation (2):
V (NO2): V (O2) = 4: 1
V (O2) = 1/4 V (NO2) = 0.25
6. Thus, the alkali solution absorbs 1 volume of NO2 and 0.25 volume of O2. Remains 0.25 of the volume of O2. The volume of the gas mixture is further reduced.
Answer: a) the volume of the mixture is reduced from 2 to 1.5 volumes; b) the volume of the mixture is reduced to 0.25 volumes.