Question

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

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Answer 1

The balanced chemical equation for the combustion of propane is

`C_3H_8+5O_2rarr3CO_2+4H_2O`.

`DeltaH=-2220` kJ/mole

Explanation:

The formula for propane is `C_3H_8`. Combustion is the reaction with oxygen to form carbon dioxide, `CO_2` and water, `H_2O` (the opposite of photosynthesis).

So our equation is

`C_3H_8+xO_2rarryCO_2+zH_2O`

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_3H_8+xO_2rarr3CO_2+zH_2O`

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_3H_8+xO_2rarr3CO_2+4H_2O`

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_3H_8+5O_2rarr3CO_2+4H_2O`

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