# Write a thermochemical equation for the combustion of propane producing 2220 kj/mol heat please?

## help

Apr 18, 2018

The balanced chemical equation for the combustion of propane is

${C}_{3} {H}_{8} + 5 {O}_{2} \rightarrow 3 C {O}_{2} + 4 {H}_{2} O$.

$\Delta H = - 2220$ kJ/mole

#### Explanation:

The formula for propane is ${C}_{3} {H}_{8}$. Combustion is the reaction with oxygen to form carbon dioxide, $C {O}_{2}$ and water, ${H}_{2} O$ (the opposite of photosynthesis).

So our equation is

${C}_{3} {H}_{8} + x {O}_{2} \rightarrow y C {O}_{2} + z {H}_{2} O$

where $x$, $y$, and $z$ are the coefficients that will balance this chemical equation.

First balance the carbon. There are 3 carbons on the left-hand side of the equation. To make 3 carbons on the right-hand side of the equation we must make $y = 3$.

${C}_{3} {H}_{8} + x {O}_{2} \rightarrow 3 C {O}_{2} + z {H}_{2} O$

Next balance the hydrogen. There are 8 hydrogens on the left-hand side of the equation. In order to make 8 hydrogens on the right-hand side of the equation we must make $z = 4$.

${C}_{3} {H}_{8} + x {O}_{2} \rightarrow 3 C {O}_{2} + 4 {H}_{2} O$

Finally balance the oxygen. There are now 10 oxygens on the right-hand side of the equation. In order to make 10 oxygens on the left-hand side of the equation we must make $x = 5$.

${C}_{3} {H}_{8} + 5 {O}_{2} \rightarrow 3 C {O}_{2} + 4 {H}_{2} O$

This is our balanced chemical equation for the combustion of propane. Because this reaction produces heat, the change in enthalpy is negative ($\Delta H = - 2220$ kJ/mole)