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

Mar 6, 2017

${C}_{5} {H}_{12} + 8 {O}_{2} \rightarrow 5 C {O}_{2} + 6 {H}_{2} O + \Delta$

#### 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:

$\left(i\right)$ $\text{Balance the carbons as carbon dioxide}$

$\left(i i\right)$ $\text{Then balance the hydrogens as water}$

$\left(i i i\right)$ $\text{And then balance the oxygens on the LHS} .$

Mar 6, 2017

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

${\text{C"_5"H"_12 + color(purple)("8")"O}}_{2}$$\rightarrow$color(red)("5")"CO"_2 + color(green)("6")"H"_2"O" + "heat"

#### Explanation:

Balance the equation:

${\text{C"_5"H"_12 + "O}}_{2}$$\rightarrow$$\text{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 ${\text{CO}}_{2}$. 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 $\text{H"_2"O}$.

${\text{C"_5"H"_12 + "O}}_{2}$$\rightarrow$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 $\text{O"_2}$.

${\text{C"_5"H"_12 + color(purple)("8")"O}}_{2}$$\rightarrow$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: $\text{5 C atoms}$, $\text{12 H atoms}$, $\text{16 O atoms}$

Right Side: $\text{5 C atoms}$, $\text{12 H atoms}$, $\text{16 O atoms}$