Here's how you approach a combustion reaction problem.

C(3)H(8) + O(2) = CO(2) + H(2)O

Determine the molecule with the most atoms. In this case it is the C(3)H(8). Put a coefficient of "1" in front of it. This will be our base molecule which we will use to balance all the atoms.

C(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 CO

**1**C(3)H(8) + O(2) = **3**CO(2) + H(2)O. Carbons are now balanced. Next we use our base molecule C(3)H(8) to determine the number of H atoms it contains. It has 8. 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.

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. We need four (4) H(2)O molecules to get 8 H atoms.

**1**C(3)H(8) + O(2) = **3**CO(2) + **4**H(2)O. 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.

That's all we can do with our base molecule, but we still need to balance the oxygen (O) atoms. Currently, we have 2 O atoms on the left and a total of 10 O atoms on the right (3x2 = 6 O atoms from the 3 CO(2) molecules and 4 O atoms from the 4 H(2)O molecules).
We need to find a coefficient for the O(2) that will yield 10 O atoms to match the 10 O atoms on the right side of the equation. That's easy. Use a coefficient of 5 in front of the O(2).

5 x 2 = 10

**1**C(3)H(8) + **5** O(2) = **3**CO(2) +
**4**H(2)O

**Left side Right side**

C= 2 x 3 = 6 C = 6 x 1 = 6

H= 2 x 8 = 16 H = 8 x 2 = 16

O = 5 x 2 = 10 O = (3 x 2) + (4 x 1) = 10

For your first problem, C(8)H(18) + O(2) = CO + H(2)O, it works in a similar way:

Choose C(8)H(18), place a coefficient of "1" in front of it.

**1**C(8)H(18) + O(2) = CO(2) + H(2)O; balance the C atoms - we need 8 on both sides

**1**C(8)H(18) + O(2) = **8**CO(2) + H(2)O; balance the H atoms - we need 18 on both sides

**1**C(8)H(18) + O(2) = **8**CO(2) + **9**H(2)O; add up the oxygens on the left and compare with the number on the right side of the equation. We have 2 O atoms on the left and a total of 25 O atoms on the right
[(8 x 2 = 16 from the CO molecules) + (9 x 1 = 9 from the H(2)O molecules)]. If we have 25 O atoms on the right, we need 25 on the left.

How do we do this? We use a coefficient of 25/2 on the left since 25/2 x 2 = 25.

**1**C(8)H(18) + **25/2** O(2) = **8**CO(2) +
**9**H(2)O

We never leave fractions as coefficients, so we multiply the entire equation by 2

**2**C(8)H(18) + **25** O(2) = **16**CO +
**18**H(2)O

For the: Pb(OH)(2) + HCl = H(2)O + PbCl(2), rewrite water [H(2)O] as H-OH

Pb(OH)(2) + HCl = H-OH + PbCl(2). Now balance the equation:

**1**Pb(OH)(2) + HCl = H-OH + PbCl(2); 1 Pb on the left, 1 Pb on the right

**1**Pb(OH)(2) + HCl = **2**H-OH + PbCl(2); 1x2 =2 OH's on the left and 2x1=2 OH's on the right. Next, balance the Cl atoms.

**1**Pb(OH)(2) + **2**HCl = **2**H-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.

We can now rewite the balanced equation with H-OH as H(2)O

Pb(OH)(2) + 2HCl = 2H(2)O + PbCl(2);

## Comments

Thanks

In the equation below that one. How does that 5 get there before the O?

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.

In the equation: 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. This means you need to add a coefficient of 3 to the product (right) side to make both sides equal. So far it should look like this: 1 C(3)H(8) + O(2) = 3 CO(2) + H(2)O. Next, we notice the 8 H's in the reactants, and add the coefficient of 4 to the water in the product side. Now it looks like this: 1 C(3)H(8) + O(2) = 3 CO(2) + 4 H(2)O. Now, count the O's on the product side: (3 x 2) + (4 x 1) = 10. 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. This gives us (5 x 2) = 10 O's on the product side. Now simply 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. Hope that helped clear things up a bit for you. Good luck!

-David

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