Balance the reaction for the combustion of pentane: ?C5H12+?O2→?CO2+?H2O Enter the four coefficients in order, separated by commas (e. G., 1,2,3,4), where 1 indicates the absence of a coefficient.

The four coefficients which balance the reaction for the combustion of pentane are; 1, 8, 5, 6.

The reaction involving the combustion of pentane to yield Carbon(IV)oxide, CO2 and Water, H2O can be balanced as follows;

To check for the authenticity of the coefficients;

Hence, the chemical equation is balanced with coefficients; 1, 8, 5, 6 respectively.

Read more:

link

The four coefficients in order, separated by commas are 1, 8, 5, 6

Explanation:

We count the atoms in order to balance this combustion reaction. In combustion reactions, the products are always water and carbon dioxide.

C₅H₁₂  +  ?O₂→  ?CO₂ + ?H₂O

We have 12 hydrogen in right side and we can balance with 6 in the left side. But the number of oxygen is odd. We add 2 in the  right side, so we have 24 H, and in the product side we add a 12.

As we add 2 in the C₅H₁₂, we have 10 C, so we must add 10 to the CO₂ in the product side.

Let's count the oxygens: 20 from the CO₂ + 12 from the water = 32.

We add 16 in the reactant side. Balanced equation is:

2C₅H₁₂  + 16O₂→  10CO₂ + 12H₂O

We also can divide by /2 in order to have the lowest stoichiometry

C₅H₁₂  + 8O₂→  5CO₂ + 6H₂O