What type of reaction is ch4+O2

What type of reaction is ch4+O2

The oxidation reaction, i.e. oxygen oxidizes both carbon and hydrogen to form CO2 and H2O. The changes of the valence of carbon and hydrogen in these two substances and the valence of carbon and hydrogen in methane are observed,

Calculation of organic chemistry (reaction of CH4 and O2)? A certain amount of CH4 and a certain amount of O2 are completely reacted to obtain a total mass of 49.6 g of CO2,CO,H2O (g). After passing through a sufficient amount of H2SO4, the washing bottle is increased by 25.2 g, and the volume of o2 is required for complete combustion of ch4. (STP)

CH4+O2==CO2+CO+H2O This is an equation for the complete reaction of a certain amount of CH4 with a certain amount of O2. For the time being, without balancing, we can find that all the hydrogen elements in CH4 are in H2O, so we only need to know how much water there is. One mole of CH4 is converted into two moles of water.
After the sufficient amount of H2SO4 is passed, the weight of the gas washing bottle is increased by 25.2g, and it can be known that the mass of water is 25.2g. Because the three gases pass through the concentrated sulfuric acid, only water is absorbed. The molar mass of 25.2g H2O is not difficult to find.
N (H2O)= m/M =25.2/18=1.4 mol, so n (CH4)=1/2 n (H2O)=0.7 mol
Now that the molar amount of CH4 is known, the amount of oxygen required for complete combustion can be known from the equation for complete combustion of CH4
CH4+2O2==CO2+2H2O
The mass of water produced by the complete combustion of 2.8 moles of CH4 remains unchanged, because the hydrogen element is completely turned into water.
The molar amount of CO2 generated by 2.8 moles of CH4 is n (CO2)= n (CH4)(conservation of carbon elements)
Therefore, the mass of CO2 is m (CO2)= n (CO2)*M (CO2)=0.7*44=30.8 g
The total mass of the product of complete combustion with sufficient oxygen is m = m (CO2)+ m (H2O)=30.8+25.2=56 g
The difference in quality between complete combustion and insufficient oxygen is the quality of oxygen that still needs to be introduced
56-49.6=6.4Gn (O2)= m (O2)/ M (O2)=6.4/32=0.2mol
V (O2)=0.2*22.4=4.48
, The complete combustion of ch4 requires another 4.48 L volume of O2