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How does these chemical equation balance, step by step? Why do they balance this way?

I already know how they balnace I just don't understand how they balance.

why does the O(2*) balance to 25? 2 C(8*)H(18*) + 25 0(2) yields 16 CO(2*) + 18 H(2*)O

These ones have me stumped as well

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

Why does this equation balance this way? 

1 Pb(OH)(2) + 2 HCl = 2 H(2)O + 1 PbCl(2) 

Pb(OH)(2)  + HCl  + H(2)O + PbCl(2) 

As I look at this equation, I am immediatly stumped because of the (OH). It kind of looks like it affects the whole equation, but I don't know why or know.  

Please explain to me, in depth, why and how these equations balance this way 

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George C. | Humboldt State and Georgetown graduateHumboldt State and Georgetown graduate
5.0 5.0 (2 lesson ratings) (2)

In your first equation you have products 16 CO(2) +18 H(2)O.  The total number of oxygen atoms is:

(16) O(2) from C(O)2, or 32 O, and 18 O from H(2)O. 32+18=50.  Oxygen as a reactant has a subscript of 2 in C(O)2.  So divide by 2 and you get 25.

Do likewise in the remaining systems.







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!


Freddie P. | Tutoring in General Chemistry, pre-Algebra, Algebra I and II, ReadingTutoring in General Chemistry, pre-Algeb...
4.9 4.9 (199 lesson ratings) (199)

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

1C(3)H(8)  +  O(2)  =  3CO(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.

1C(3)H(8)  +  O(2)  =  3CO(2)  +  4H(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

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

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

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

1C(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.  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.

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

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

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

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

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

1Pb(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.

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

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


A correction to the atom balance in the first equation. 1C(3)H(8) + 5 O(2) = 3CO(2) + 4H(2)O C=1x3=3 C=3x1=3 H=1x8=8 H=4x2=8 O=5x2=10 O=(3x2)+(4x1)=10
Robert C. | Dr. Robert can help you with Math and ScienceDr. Robert can help you with Math and Sc...
5.0 5.0 (44 lesson ratings) (44)

Hi Andrew.

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

Some of the numbers, called coefficients, are determined from experiments. The others are filled it to make it balance. Consider this reaction:

1 C3H8 + 5 O2 = 3 CO2 + 4 H2O

At some point this reaction may have been studied and the C3H8 formula was not known. What could be determined is that when 1 mole of the unknown was burned, it took 5 moles of O2 gas to burn it completely and that the only products were CO2 and steam (H2O). This would suggest that the unknown had only C, H, and O in it. Further testing might have revealed that it only had C and H in it.

Also, if we didn't know how much O2 was needed, we could measure the amount of CO2 and H2O that is produced and realize that for each mole of the unknown (C3H8) that is burned, 3 moles of CO2 and 4 moles of H2O are produced. Then the 5 could be deduced because it is the only number you can put in front of the O2 that balances the equation.

The principle at work is the conservation of mass. The chemical reaction does not create or destroy any atoms or elements. What you start with must be what you finish with in terms of matter.


As far as the OH term. When you see a formula like Pb(OH)2 , it means that the O and the H are to be considered a single unit. The 2 means that 2 of these units are attached to the lead (Pb) atom. The 2 applies to everything inside the parentheses. So, when you do the math, a single Pb(OH)2 is 1 Pb, 2 O, and 2 H.