what is the balanced equation when an incomplete combustion of fuel in a butane (c4h10) gas stove produces carbon monoxide and water vapors

Answer 1

For this problem, we want to first write out our reactants and our products. Butane and O₂ (O₂ will always be a reactant for a combustion reaction) are our reactants while our products are carbon monoxide and water

C₄H₁₀ + O₂ ----> CO₂ + H₂O

Now that we have our equation, we need to balance it. To do this I like to write out each element and how many we have on each side of the equation.

C₄H₁₀ + O₂ ----> CO₂ + H₂O

C: 4 | C: 1

H: 10 | H: 2

O: 2 | O: 3

Now, when balancing combustion reactions particularly, you should balance oxygen last. The reason for this is that we see it in two molecules on the right side of the equations so it is hard to determine which one should be increased to balance the left side. Thus, I start with carbon, then hydrogen, and lastly oxygen.

First carbon (it changes also the number of oxygens here don't forget- now 8 from CO₂ vs the 2 before):

C₄H₁₀ + O₂ ----> 4 CO₂ + H₂O

C: 4 | C: 4

H: 10 | H: 2

O: 2 | O: 9

Now hydrogen:

C₄H₁₀ + O₂ ----> 4 CO₂ + 5 H₂O

C: 4 | C: 4

H: 10 | H: 10

O: 2 | O: 13

Lastly oxygen:

C₄H₁₀ + 6.5 O₂ ----> 4 CO₂ + 5 H₂O

C: 4 | C: 4

H: 10 | H: 10

O: 13 | O: 13

HOWEVER, we cannot have decimals or fractions for the coefficient (cannot have 6.5). To get rid of this .5 we would need to double each molecule's coefficients.

Removing the decimal coefficient :

2 C₄H₁₀ + 13 O₂ ----> 8 CO₂ + 10 H₂O

C: 8 | C: 8

H: 20 | H: 20

O: 26 | O: 26

Therefore the final answer should be (don't forget phases!!!)

2 C₄H_{10 (g)} + 13 O_{2 (g)} ----> 8 CO_{2 (g)} + 10 H₂O _(g)

I hope this was easy to follow and allows to approach balancing equation questions