What you start with must be what you finish with in terms of matter.ДЄs far as the OH term. The chemical reaction does not create or destroy any atoms or elements. The principle at work is the conservation of mass. Then the 5 could be deduced because it is the only number you can put in front of the O 2 that balances the equation. Further testing might have revealed that it only had C and H in it.ДЄlso, if we didn't know how much O 2 was needed, we could measure the amount of CO 2 and H 2O that is produced and realize that for each mole of the unknown (C 3H 8) that is burned, 3 moles of CO 2 and 4 moles of H 2O are produced. This would suggest that the unknown had only C, H, and O in it. What could be determined is that when 1 mole of the unknown was burned, it took 5 moles of O 2 gas to burn it completely and that the only products were CO 2 and steam (H 2O). Consider this reaction:ДЄt some point this reaction may have been studied and the C 3H 8 formula was not known. The others are filled it to make it balance. Some of the numbers, called coefficients, are determined from experiments. You asked 'In the equation below that one. We can now rewite the balanced equation with H-OH as H(2)O Next, balance the Cl atoms.ДЎPb(OH)(2) + 2HCl = 2H-OH + PbCl(2) Now we have 2 Cl atoms on the left, 2 Cl atoms on the right 2 H atoms on the left, 2H atoms on the right 2 OH groups on the left and 2 OH groups on the right 1 Pb on the left, 1 Pb on the right. Now balance the equation:ДЎPb(OH)(2) + HCl = H-OH + PbCl(2) 1 Pb on the left, 1 Pb on the rightДЎPb(OH)(2) + HCl = 2H-OH + PbCl(2) 1x2 =2 OH's on the left and 2x1=2 OH's on the right. We never leave fractions as coefficients, so we multiply the entire equation by 2ДЇor the: Pb(OH)(2) + HCl = H(2)O + PbCl(2), rewrite waterВ В as H-OH How do we do this? We use a coefficient of 25/2 on the left since 25/2 x 2 = 25. If we have 25 O atoms on the right, we need 25 on the left. We have 2 O atoms on the left and a total of 25 O atoms on the right. O =В 5 x 2 = 10 O = (3 x 2) + (4 x 1) = 10ДЇor your first problem, C(8)H(18) + O(2) = CO + H(2)O, it worksВ in a similar way:Д¬hoose C(8)H(18), place a coefficient of '1' in front of it.ДЎC(8)H(18) + O(2) = CO(2) + H(2)O balance the C atoms - we need 8 on both sidesДЎC(8)H(18) + O(2) = 8CO(2) + H(2)O balance the H atoms - we need 18 on both sidesДЎC(8)H(18) + O(2) = 8CO(2) + 9H(2)O add up the oxygens on the left and compare with the number on the right side of the equation. Use a coefficient of 5 in front of the O(2). We need to find a coefficient for the O(2) that will yieldВ 10 O atoms to match theВ 10 O atoms on the right side of the equation. Currently, we have 2 O atoms on the left and a total ofВ 10 O atoms on the right (3x2 = 6В O atoms from the 3 CO(2) molecules and 4 O atoms from the 4 H(2)O molecules). That's all we can do with our base molecule, but we still need to balance the oxygen (O) atoms. Now we have 3 carbons on the left side and 3 carbons on the right side of the equation we also have 8 hydrogens on the left and 8 (4 x 2) hydrogens on the right. We need to have 8 in order to balance the H atoms, so we place a coeffidient of 4 in front of the H(2)O. On the right side of the equation we notice the only H atoms are in H(2)O, and there are 2 H atoms there. Next we use our base molecule C(3)H(8) to determine the number of H atoms it contains. This will be our base molecule which we will use to balance all the atoms.Д¬(3)H(8) has 3 carbons while CO has only 1 carbon so, we need to place a coefficient of 3 in front of the COДЎC(3)H(8) + O(2) = 3CO(2) + H(2)O. Here's how you approach a combustion reaction problem.Дetermine the molecule with the most atoms. Hope that helped clear things up a bit for you. Go back and double check your quantities for C, H, and O individually to make sure it is balanced correctly, and you should be good to go. This gives us (5 x 2) = 10 O's on the product side. Since the reactant side has Oxygen as a diatomic molecule (meaning there is a subscript of 2 and therefore 2 times whatever coefficient we add), we need to place the coefficient 5 in front. Now, count the O's on the product side: (3 x 2) + (4 x 1)

Next, we notice the 8 H's in the reactants, and add the coefficient of 4 to the water in the product side.

So far it should look like this: 1В C(3)H(8) This means you need to add a coefficient of 3 to the product (right) side to make both sides equal. In the equation:В C(3)H(8) + O(2) = CO(2) + H(2)O, you want to start by counting 3 C's on the reactant (left) side. For combustion reactions involving a simple hydrocarbon (like the first two equations you posted), you want to start by balancing the carbon, then the hydrogen, and finally the oxygen.

Organic chemical equation balancer