What is the balanced equation for combustions of propane gas that yields carbon dioxide and water?

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What is the balanced equation for combustions of propane gas that yields carbon dioxide and water?

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Answer 1 · 2014-05-15T17:25:56

A combustion reaction is a reaction between a hydrocarbon burned in oxygen to form carbon dioxide and water.

A hydrocarbon is a molecule composed of carbon and hydrogen in this case propane. The pro- prefix tells us we have 3 carbon.
For hydrocarbons that end in the suffix -ane the formula is $C_{n} H_{2 n + 2}$ $C_nH_(2n+2)$ . This means we need 8 hydrogen, making the formula $C_{3} H_{8}$ $C_3H_8$ .

Now we can format the basic reaction.

$C_{3} H_{8} + O_{2} \to C O_{2} + H_{2} O$ $C_3H_8 + O_2 -> CO_2 + H_2O$

A reminder that oxygen is a diatomic molecule as a gas, $O_{2}$ $O_2$ .

Begin by balancing the hydrogen by adding a coefficient of 4 in from of the water.

$C_{3} H_{8} + O_{2} \to C O_{2} + 4 H_{2} O$ $C_3H_8 + O_2 -> CO_2 + 4H_2O$

This balances the hydrogen at 8. Now add a a coefficient of 3 in from of the carbon dioxide.

$C_{3} H_{8} + O_{2} \to 3 C O_{2} + 4 H_{2} O$ $C_3H_8 + O_2 -> 3CO_2 + 4H_2O$

This balances the carbon at 3. We now have 10 oxygen on the on the product side, 6 from $C O_{2}$ $CO_2$ and 4 from $H_{2} O$ $H_2O$ . We need to add a coefficient of 5 in front of the oxygen to balance the oxygen at 10.

$C_{3} H_{8} + 5 O_{2} \to 3 C O_{2} + 4 H_{2} O$ $C_3H_8 + 5O_2 -> 3CO_2 + 4H_2O$

The combustion reaction is now balanced.

I hope this was helpful.
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What is the balanced equation for combustions of propane gas that yields carbon dioxide and water?

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