The coefficients in the order CH₄, O₂, CO₂, and H₂O, are 1,2,1,2
The equation becomes:
CH₄ (g) + 2O₂ (g) -> CO₂ (g) + 2H₂0 (g)
<h3><em>Further explanation</em></h3>
Complete combustion of hydrocarbon with oxygen will be obtained by CO₂ and H₂O compounds.
If O is insufficient there will be incomplete combustion produced by CO and H and O
Hydrocarbon combustion reactions (specifically alkanes)
![\large {\boxed {\bold {C_nH _ (_2_n _ + _ 2_) + \frac {3n + 1} {2} O_2 ----> nCO_2 + (n + 1) H_2O}}}](https://tex.z-dn.net/?f=%5Clarge%20%7B%5Cboxed%20%7B%5Cbold%20%7BC_nH%20_%20%28_2_n%20_%20%2B%20_%202_%29%20%2B%20%5Cfrac%20%7B3n%20%2B%201%7D%20%7B2%7D%20O_2%20----%3E%20nCO_2%20%2B%20%28n%20%2B%201%29%20H_2O%7D%7D%7D)
Equalization of chemical reaction equations can be done using variables.
Steps in equalizing the reaction equation:
- 1. gives a coefficient on substances involved in the equation of reaction such as a, b, or c etc.
- 2. make an equation based on the similarity of the number of atoms where the number of atoms = coefficient × index between reactant and product
- 3. Select the coefficient of the substance with the most complex chemical formula equal to 1
For gas combustion reaction which is a reaction of hydrocarbons with oxygen produces CO₂ and H₂O (water vapor). can use steps:
Balancing C atoms, H and last atoms O atoms
Methana combustion reaction
CH₄ (g) + O₂ (g) -> CO₂ (g) + H₂0 (g)
We give the most complex compounds, namely Methane, we give the number 1
So the reaction becomes
CH₄ (g) + aO₂ (g) -> bCO₂ (g) + cH₂0 (g)
C atom on the left 1, right b, so b = 1
H atom on the left 4, right 2c, so 2c = 4 ---> c = 2
Atom O on the left 2a, right 2b + c, so 2a = 2b + c
2a = 2.1 + 2
2a = 4
a = 2
The equation becomes:
CH₄ (g) + 2O₂ (g) -> CO₂ (g) + 2H₂0 (g)
<h3><em>Learn more</em></h3>
the combustion of octane in gasoline
brainly.com/question/8175791
brainly.com/question/897044
hydrogen and excess oxygen
brainly.com/question/1405182
Keywords: methane, combustion, hydrocarbons, equalization of reaction equations