Question

How do you balance this equation: `?C_5H_12 + ?O_2 → ?CO_2 + ?H_2O + heat`?

Answer 1

`C_5H_12 + 8O_2 rarr 5CO_2 +6H_2O + Delta`

Explanation:

Well is it stoichiometrically balanced? Garbage out must equal garbage in if it is to be a proper representation of chemical reality. And not only is it balanced with respect to mass and charge, it is also balanced in terms of energy transfer. A given quantity of pentane results in a precise quantity of energy upon complete combustion.

As to how to do it, the usual rigmarole is to:

`(i)` `"Balance the carbons as carbon dioxide"`

`(ii)` `"Then balance the hydrogens as water"`

`(iii)` `"And then balance the oxygens on the LHS".`

See [here for another example.](https://socratic.org/questions/what-is-the-product-of-c6h5oh-reacting-with-oxygen-gas-and-how-do-i-balance-it?answerEditSuccess=1)

Answer 2

1. Balance the carbon atoms.
2. Balance the hydrogen atoms.
3. Balance the oxygen atoms.

`"C"_5"H"_12 + color(purple)("8")"O"_2` `rarr` `color(red)("5")"CO"_2 + color(green)("6")"H"_2"O" + "heat"`

Explanation:

Balance the equation:

`"C"_5"H"_12 + "O"_2` `rarr` `"CO"_2 + "H"_2"O" + "heat"`

Due to the law of conservation of mass/matter , the number of atoms for each element must be the same on both sides. When balancing a chemical equation, the chemical formulas are never changed , which means subscripts are never changed. What can change is the amount of each reactant and product. The amount is indicated by a coefficient in front of a formula. The coefficient is multiplied by the subscripts of the elements in the formula.

The strategy for balancing combustion reactions is:

1. Balance the carbon atoms.
2. Balance the hydrogen atoms.
3. Balance the oxygen atoms.

Carbon and Hydrogen

There are 5 C atoms on the left and 1 C atom on the right. Place a coefficient of `5` in front of the `"CO"_2`. There are 12 H atoms on the left side, and 2 H atoms on the right. Place a coefficient of `6` in front of the `"H"_2"O"`.

`"C"_5"H"_12 + "O"_2` `rarr` `color(red)("5")"CO"_2 + color(green)("6")"H"_2"O" + "heat"`

Oxygen

There are 2 O atoms on the left and 16 O atoms on the right. Place a coefficient of `8` in front of the `"O"_2"`.

`"C"_5"H"_12 + color(purple)("8")"O"_2` `rarr` `color(red)("5")"CO"_2 + color(green)("6")"H"_2"O" + "heat"`

Check the numbers of atoms of each element on each side.

Left Side: `"5 C atoms"`, `"12 H atoms"`, `"16 O atoms"`

Right Side: `"5 C atoms"`, `"12 H atoms"`, `"16 O atoms"`