Q:

# What is the balanced equation for the combustion of butane?

A:

Two molecules of butane combine with 13 molecules of oxygen to yield eight molecules of carbon dioxide and ten molecules of water. The balancing process requires carefully adjusting the numbers of one reactant or product at a time.

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Credit: Joe Haupt CC-BY-2.0

Begin by writing the unbalanced equation for butane's combustion, C4H10 plus O2 yields CO2 plus H2O. The left side has four carbon, two oxygen and ten hydrogen atoms, while the right side has just one hydrogen atom along with one carbon and two oxygen.

Start by multiplying the CO2 in the products by four to match the four carbons in butane on the reactant side. Next, multiply the H2O in the products by five, resulting in ten hydrogen atoms in both the reactants and products. At this point, there are 13 oxygen atoms in the products, eight in the carbon dioxide and five in the water. Since the oxygen gas in the reactants is a diatomic molecule, 6.5 units of oxygen gas balances out the 13 oxygen atoms in the products. For clarity, double everything in the reactants and products; 2 C4H10 plus 13 O2 reacts to give 8 CO2 plus 10 H2O.

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