Reaction+Funf-+Combustion


 * COMBUSTION **

Combustion occurs when a substance combines with oxygen to produce at least one oxide. It will always involve Carbon, Hydrogen, and Oxygen, and will always result in at least 1 water molecule.

Generically, a decomposition equation looks like this: A+O 2  à CO 2 + H 2 O

An actual equation could be: Dicarbon Hydrogen gas and Dioxide form Carbon Dioxide and Water In order to change this into a formula, you must first know the symbols for each element in the equation Water=H 2 O Hydrogen=H Oxygen=O Carbon=C The “di” in dicarbon and dioxide allude to there being a subscript of 2 next to the elements symbol. Also, hydrogen being a gas makes the subscript next to the H in the reactants also 2. The word “form” in the current equation indicates an arrow (  à ). Thus far we know: C 2 H 2 +O 2  à CO 2 + H 2 O  FINALLY…BALANCING!! In order to balance this equation (like any other equation) you must figure out how much of each element there are on each side of the equation. On the reactants (left) side, there are two Carbons, Hydrogens, and Oxygens. However, on the products (right) side, there are two hydrogens, one carbon, and 3 oxygens! If you would like to put this into a nifty little table to organize the thoughts more clearly, you can do so. An example pertaining to this equation would be as follows: ___C___ __ 2 __C__ 1 __ ___H___ you fill in the blanks to make it: __ 2 H 2 __ ___O___ __ 2 O 3__ You will know the equation is balanced when all those numbers match the one across it. So far, the H’s are already balanced, but the O’s and C’s need some work. Let’s start with O’s. There are two on one side, and 3 on the other. One of these does not go into the other, so we will have to multiply them both. You will probably have to mess around with a couple of different numbers before getting the right ones, but considering I have the answers right next to me, I will tell you that in order to make the O’s balanced, the equation must look like this: __C__ 2 H 2 + 5 O 2  à 4 CO 2 + 2 H 2 O  This gives you 10 oxygens on each side! I’m glad you noticed that. You see, there is still more balancing to be done. As the equation stands, there are 10 oxygens on each side, but now the H’s and C’s are uneven. Your table currently looks like this: 2 __ C __ 4 2 H 4 10 O 10 Luckily, in this situation, all we need to do is double the C 2 H 2 in order to balance the equation. So, you place a 2 where that space in the equation. Now, your table looks like this: 4 __ C __ 4 4 H 4 10 O 10 And your final equation looks like this: and everyone is happy.  J
 * But, Lauren, there is still a space in the equation; I thought it was all balanced!?!? **
 * 2C **** 2 **** H **** 2 **** + 5 O **** 2 ** ** à **** 4 CO **** 2 **** + 2 H **** 2 **** O **

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