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Write and balance the equation for the combustion of C6H14.

Â When we have a combustion reaction, what that means is that we're adding oxygen.

Â And for our hydrocarbons, our only products are going to be CO2 and water.

Â So, I'm going to start by writing the basics of our reaction.

Â So, we have C6H14 plus 02

Â yields C02 plus water.

Â And because this isn't going to start out balanced, I'm leaving some room for

Â my coefficients as I go through and balance the equation.

Â Now, I'm going to create a list of my elements so

Â that I can basically create a tally.

Â C, H and O of how many carbons, hydrogens, and

Â oxygens I have on both sides of the equation.

Â So, on the left side, I see I have 6 carbons, 14 hydrogens, and 2 oxygens.

Â On the right side, I have 1 carbon, 2 hydrogens and 3 oxygens.

Â So, now what I need to do is start putting in coefficients and adjusting my

Â totals until I get all the coefficients I need to have a balanced chemical equation.

Â So, I'm going to start by putting a six in front of the CO2,

Â so that I can balance the carbons.

Â So, that changes from one carbon to six carbons on the right, but

Â notice that it also changes the number of oxygens.

Â Now I have 12 plus one, or 13 oxygens on the right side of the equation.

Â So, now my carbons are balanced, now I'm going to look at the hydrogens, and

Â in order to balance the hydrogens, I need a 7 on the right side of the equation.

Â So, now I have 14 hydrogens, and that also changes the number of oxygens, so,

Â now I have 12 oxygens plus seven, or 19 oxygens.

Â Now at this point, both my carbons and my hydrogens are balanced.

Â It's only the oxygens that are, the not balanced.

Â And when I look at the left side,

Â what I notice is that I have oxygen all by itself.

Â So what I'm going to do, is for a moment, I'm going to pretend.

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Then I would have a balanced equation.

Â But that's assuming that the two is not there.

Â But I could see that I have the right number of everything.

Â But I do have the to in there, so

Â what I have to do is actually change that coefficient to 19 halves, and

Â return that two into the equation.

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So, now I actually do have a balanced equation but I still have

Â this fractional number in my coefficients which I don't want to leave in there.

Â So what I can do, is I can multiply everything through by two.

Â And when I do that I get a 2 here, I get a 19,

Â in front of the oxygen, I get 12 in

Â front of CO2 and I get 14 in front of H2O.

Â Now I can go back and check my count, I see that I have 12 carbons on the left,

Â I have 28 hydrogens, and I have 38 oxygens.

Â On the right side of my equation, I have 12 carbons,

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Now, I have a balanced chemical equation with integers as coefficients and

Â I see that that's the lowest possible set of integers because I

Â cannot divide it by any number and still get whole numbers.

Â So, it looks a little odd to see values that are this large but

Â we do see this when we start looking at larger reactants,

Â particularly when we look at combustion reactions.

Â