Discussion and Twenty Examples

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Balance redox equations by sight |

Chemical equations usually do not come already balanced. Making sure they are balanced must be done before the equation can be used in any chemically meaningful way.

All chemical calculations you will see in other units must be done with a balanced equation.

**IMPORTANT DEFINITION:** A balanced equation has equal numbers of each type of atom on each side of the equation.

The Law of Conservation of Mass is the rationale for balancing a chemical equation. The law was discovered by Antoine Laurent Lavoisier (1743-94) and this is his formulation of it, translated into English in 1790 from the *Traité élémentaire de Chimie* (which was published in 1789):

"We may lay it down as an incontestible axiom, that, in all the operations of art and nature, nothing is created; an equal quantity of matter exists both before and after the experiment; the quality and quantity of the elements remain precisely the same; and nothing takes place beyond changes and modifications in the combination of these elements."

A less wordy way to say that is this:

"Matter is neither created nor destroyed."

Therefore, we must finish our chemical reaction with as many atoms of each element as when we started.

**Example #1:** Balance the following equation: H_{2} + O_{2} ---> H_{2}O

It is an unbalanced equation (sometimes also called a skeleton equation). This means that there are UNEQUAL numbers at least one atom on each side of the arrow. By the way, a skeleton equation is not wrong, it just hasn't been balanced yet. Presenting it as being balanced would be wrong.

In the example equation, there are two atoms of hydrogen on each side, BUT there are two atoms of oxygen on the left side and only one on the right side.

**Remember this:** A balanced equation MUST have EQUAL numbers of EACH type of atom on BOTH sides of the arrow.

An equation is balanced by changing coefficients in a somewhat trial-and-error fashion. It is important to note that only the coefficients can be changed, NEVER a subscript.

**Important point:** the coefficient times the subscript gives the total number of atoms.

Four examples before balancing the equation.

(a) 2H_{2}---> there are 2 x 2 atoms of hydrogen (a total of 4).(b) 2H

_{2}O ---> there are 2 x 2 atoms of hydrogen (a total of 4) and 2 x 1 atoms of oxygen (a total of 2).(c) 2(NH

_{4})_{2}S ---> there are 2 x 1 x 2 atoms of nitrogen (a total of 4), there are 2 x 4 x 2 atoms of hydrogen (a total of 16), and 2 x 1 atoms of sulfur (a total of 2).(d) 3Ca(NO

_{3})_{2}(just the oxygens) ---> There are 18. The 3 on the nitrate times 2 outside the parenthesis equals 6 oxygen in one formula unit. The coefficient of three times the 6 gives the final answer of 18.

Now, back to balancing the example equation:

H_{2}+ O_{2}---> H_{2}O

The hydrogen are balanced, but the oxygens are not. We have to get both balanced. We put a two in front of the water and this balances the oxygen.

H_{2}+ O_{2}---> 2H_{2}O

However, this causes the hydrogen to become unbalanced. To fix this, we place a two in front of the hydrogen on the left side.

2H_{2}+ O_{2}---> 2H_{2}O

This balances the equation.

Four points to make:

1) You CANNOT change a subscript.

You cannot change the oxygen's subscript in water from one to two, as in:H_{2}+ O_{2}---> H_{2}O_{2}True, this is a balanced equation, but you have changed the substances in it. H

_{2}O_{2}is a completely different substance from H_{2}O. So, it's not the answer to the question that was asked.

2) You CANNOT place a coefficient in the middle of a formula.

The coefficient goes at the beginning of a formula, not in the middle, as in this example:H_{2}+ O_{2}---> H_{2}2OWater only comes as H

_{2}O and you can only use whole formula units of it.Note: I actually had a student do this. I was gentle in my correction of his mistake.

3) Make sure that your final set of coefficients are all whole numbers with no common factors other than one. For example, this equation is balanced:

4H_{2}+ 2O_{2}---> 4H_{2}OHowever, all the coefficients have the common factor of two. Divide through to eliminate common factors like this.

Technically, the equation just above is balanced, but only if you ignore the "no common factors other than one" portion. The correct answer has all common factors greater than one removed. If you were to answer a test question balanced as above, you will probably only get partial credit, if that.

By the way, there are uses for balanced equations where the coefficients share a common factor of 2, 3, or more. Those uses will show up in other units that are taught through the school year. However, be aware that common factors greater than one are banned in the balancing unit.

4) NO fractions allowed in the final answer, only whole numbers. For example:

H_{2}+^{1}⁄_{2}O_{2}---> H_{2}Ois an allowable step along the way to the answer, but it is not the final answer.

The use of fractions in balancing is a powerful tool. Look for it in the solved examples.

From here on out, I will simply give the equation to be balanced.

**Example #2:** H_{2} + Cl_{2} ---> HCl

Remember that the rule is: A balanced equation MUST have EQUAL numbers of EACH type of atom on BOTH sides of the arrow.

The correctly balanced equation is:

H_{2}+ Cl_{2}---> 2HCl

Placement of a two in front of the HCl balances the hydrogen and chlorine at the same time. This happens fairly often at the end of a balancing sequence, when the placement of one coefficient balances two different elements at the same time.

This equation can be balanced using fractions:

^{1}⁄_{2}H_{2}+ Cl_{2}---> HClfollowed by:

^{1}⁄_{2}H_{2}+^{1}⁄_{2}Cl_{2}---> HCl <--- this equation is balanced, but is not the final answerMultiply through by 2 to give the final answer of:

H

_{2}+ Cl_{2}---> 2HCl

Comment: down at the bottom of the file (Bonus Example #2) is a problem that uses fractions twice within the balancing sequence for the equation. I haven't seen one like that throughout my entire career, which started well before 2020 (which is when I came across the example).

**Example #3:** O_{2} ---> O_{3}

1) Hint: think about what the least common multiple is between 2 and 3. That's right - six.

2) The LCM tells you how many of each atom will be needed. Your job is to pick coefficients that get you to the LCM.

3) The correctly balanced equation is:

3O_{2}---> 2O_{3}

4) One day, I decided to balance this equation using fractions. Here it is:

^{1}⁄_{2}O_{2}--->^{1}⁄_{3}O_{3}<--- one O on each side

5) Clear one fraction by multiplying through by 2:

O_{2}--->^{2}⁄_{3}O_{3}<--- two O on each side

6) Clear the second fraction by multiplying through by 3:

3O_{2}---> 2O_{3}<--- six O on each side

**Example #4a:** Na + H_{2}O ---> NaOH + H_{2}

In the skeleton equation as written, the Na and the O are already balanced. So, we look only at the H.

Notice that the H must come in twos on the left-hand side. That means we must have an even number of hydrogen on the right-hand side. We do this:

Na + H_{2}O ---> 2NaOH + H_{2}

to make an even number of hydrogens on the right. We then balance the H like this:

Na + 2H_{2}O ---> 2NaOH + H_{2}

Notice that the first placing of a 2 messed up the balance of the Na and the O. In addition, notice that the second placing of a 2 balances the oxygen and the hydrogen at the same time.

The last step is to balance the Na:

2Na + 2H_{2}O ---> 2NaOH + H_{2}

and it's done.

Here's another method. Go back to the skeleton equation and balance it like this:

Na + H_{2}O ---> NaOH +^{1}⁄_{2}H_{2}

What you did with ^{1}⁄_{2}H_{2} is reduce the number of hydrogens on the right-hand side from a total of 3 to a total of 2. Since everything but the hydrogen was already in balance, the equation is now balanced.

Multiply through by 2 to get the final answer (which should not have any fractions in it).

**Example #4b:** Here's another example where you reduce the amount of something in order to balance the equation:

Na_{2}O_{2}+ H_{2}O ---> NaOH + O_{2}

1) Balance the sodium:

Na_{2}O_{2}+ H_{2}O ---> 2NaOH + O_{2}

2) Notice that the hydrogen is also balanced by putting 2 in front of the NaOH. The reduction occurs with the oxygen:

Na_{2}O_{2}+ H_{2}O ---> 2NaOH +^{1}⁄_{2}O_{2}

3) You reduce the oxygen from four to three on the right-hand side:

Na_{2}O_{2}+ H_{2}O ---> 2NaOH +^{1}⁄_{2}O_{2}

4) Multiply through by 2 for the final answer.

**Example #4c:** Another reduction of amount, this time on the reactant side:

SO_{2}+ O_{2}+ H_{2}O ---> H_{2}SO_{4}

1) Notice that the sulfur and the hydrogen are already balanced. Leave them alone.

2) There are five oxygens on the left and four on the right. Trying the balance the O by using the SO_{2}, the H_{2}O, or the
H_{2}SO_{4} would affect the S balance or the H balance.

3) We look to the O_{2} to save us:

SO_{2}+^{1}⁄_{2}O_{2}+ H_{2}O ---> H_{2}SO_{4}

4) The equation is now balanced. Multiply through by 2 for the final answer.

**Example #4d:** One more equation where you reduce a coefficient in order to balance the equation:

H_{2}O_{2}---> H_{2}O + O_{2}

1) Two oxygen on the left and three on the right. Reduce!!

H_{2}O_{2}---> H_{2}O +^{1}⁄_{2}O_{2}

2) Multiply through by 2 for the final answer:

2H_{2}O_{2}---> 2H_{2}O + O_{2}

**Example #5:** Fe + CuSO_{4} --> Cu + FeSO_{4}

As you examine the equation, you see that it is already balanced. There is one Fe on each side, one Cu, one S and four oxygens (the last two can also be identified as one sulfate on each side).Nothing needs to be done to this equation. As it is written on the paper (or screen), it is balanced.

You sometimes see this presented using the phrase "balanced as written."

Also, students are sometimes confused by this type of problem (sometimes even suspicious that a trick is being played on them). In many cases, a student has seen example after example where something is done to the equation to balance it. The above example needs nothing to be done; it is already balanced as written.

Here are more examples of 'balanced as written:'

C(s) + O_{2}(g) ---> CO_{2}(g)Na

_{2}SO_{3}(aq) + H_{2}SO_{4}(aq) ---> Na_{2}SO_{4}(aq) + SO_{2}(g) + H_{2}O(ℓ)CaCl

_{2}(aq) + Pb(NO_{3})_{2}(aq) ---> Ca(NO_{3})_{2}(aq) + PbCl_{2}(s)CaO(s) + H

_{2}O(ℓ) ---> Ca(OH)_{2}(s)Hg

_{2}CO_{3}---> Hg + HgO + CO_{2}HCl + NaOH ---> NaCl + H

_{2}OHCOOH + NaOH ---> HCOONa + H

_{2}OCu + CO

_{2}---> CuO + COSometimes, a writer will use 'balanced as written' for any old equation that is already balanced (using coefficients) when presented to you. The ChemTeam prefers to use the term only for equations that do not need any balancing ever. In other words, for reactions that are balanced with only coefficients of one being used.

**Example #6:** Zn + HCl ---> ZnCl_{2} + H_{2}

Upon examining this equation, you see that there is already one Zn on each side of the equation. We will attempt to leave it alone, if at all possible, since it is already balanced.One the right side, we see two chlorines and two hydrogens, with only one of each on the left. Putting a two in front of the HCl doubles the number of chlorine and hydrogen on the left side.

This now leaves us with two chlorines and two hydrogens on each side of the arrow, making them both balanced.

Since the zinc was already balanced, the entire equation is now balanced.

Zn + 2HCl ---> ZnCl

_{2}+ H_{2}

**Example #7a:** KClO_{3} ---> KCl + O_{2}

Start by noticing the the K and the Cl are ALREADY balanced in the skeleton equation. However, the oxygen is out of balance with three on the left and two on the right.It is important to emphasize that the oxygen on the left will increase only in steps of three, while the oxygen on the right will increase only in steps of two. The question to ask yourself is "What is the least common multiple between 2 and 3?" The answer of course is six. We need six oxygens on each side of the equation. We use a two on the left side since 2 x 3 = 6 and we use a three on the right side since 3 x 2 = 6.

This causes the K and the Cl to become unbalanced, but putting a two in front of the KCl on the right side fixes that.

This problem is interesting because you focused on the oxygens first. Normally, oxygen is the last (or next-to-last) element to be balanced.

2KClO

_{3}---> 2KCl + 3O_{2}Another way to balance this equation is by using a fractional coefficient:

KClO

_{3}---> KCl +^{3}⁄_{2}O_{2}And, in one step, it's balanced. You then multiply through by 2 to get the whole number set of coefficients, the 2, 2, 3 just above.

**Example #7b:** KClO_{3} ---> KClO_{4} + KCl

What is the least common multiple between 3 and 4? That's right, 12. Watch this:4KClO

_{3}---> 3KClO_{4}+ KClAnd it's balanced. Just like that. Pretty slick, huh?

**Example #7c:** H_{3}PO_{3} ---> H_{3}PO_{4} + PH_{3}

Another example where the LCM is 12:4H

_{3}PO_{3}---> 3H_{3}PO_{4}+ PH_{3}And done.

**Example #8:** S_{8} + F_{2} ---> SF_{6}

An eight in front of the SF_{6}will balance the sulfurs.This now gives us 48 fluorines on the right-hand side, since 8 x 6 = 48. Use a 24 in front of F

_{2}since 24 x 2 also equals 48.S

_{8}+ 24F_{2}---> 8SF_{6}

**Example #9:** Fe + O_{2} ---> Fe_{2}O_{3}

In the unbalanced equation, there was only one Fe on the left and two on the right. Putting a two in front of the Fe on the left brings the irons into balance.The situation balancing the oxygen is quite common. You saw it in a previous example. This time, I'll try to lay it out in steps.

- The oxygen on the left ONLY comes in twos, while the right-hand side oxygen ONLY comes in threes.
- We have to get an equal number of oxygens. (Remember, we can only adjust the value of the coefficient. We cannot change the subscript.)
- The least common multiple between two and three is six. This means we will need six oxygens on each side of the equation.
- To get this, we put a three in front of the O
_{2}since 3 x 2 = 6 and we put a two in front of the Fe_{2}O_{3}since 2 x 3 = 6.The Fe was balanced, but has become unbalanced as a consequence of our work with the oxygen. Putting a four in front of the Fe on the left solves this.

4Fe + 3O_{2}---> 2Fe_{2}O_{3}This equation could also have been balanced using a fractional coefficient:

2Fe +^{3}⁄_{2}O_{2}---> Fe_{2}O_{3}Then, multiply through by 2 to get the coefficients for the final answer shown above.

**Example #10:** C_{2}H_{6} + O_{2} ---> CO_{2} + H_{2}O

First, balance the carbons with a two in front of the CO_{2}. Then balance the hydrogens with a three in front of the H_{2}O. This leaves the following equation:C_{2}H_{6}+ O_{2}---> 2CO_{2}+ 3H_{2}OOnly the oxygens remain to be balanced, but there is a problem. On the right side of the equation, there are seven oxygen atoms, BUT oxygen only comes in a group of two atoms on the left side. Another way to say it - with O

_{2}it is impossible to generate an ODD number of oxygen atoms.However, that is true only if you were using whole number coefficients. It is allowable to use

FRACTIONALcoefficients in the balancing process. That means I can use seven-halves as a coefficient to balance this equation, like this:C_{2}H_{6}+^{7}⁄_{2}O_{2}---> 2CO_{2}+ 3H_{2}OGenerally, the fractional coefficient is not retained in the final answer. Multiplying the coefficients through by two gets rid of the fraction. here is the final answer:

2C_{2}H_{6}+ 7O_{2}---> 4CO_{2}+ 6H_{2}OAlso, improper fractions like

^{7}⁄_{2}should be used rather than a mixed number like 3^{1}⁄_{2}. Sometimes a decimal will be used (in the example being used, it would be a 3.5). The ChemTeam tends to not use decimal values in coefficients, but there are those who prefer it. Be prepared!

**Example #11-13:** Balance these three equations using fractional coefficients:

S_{8}+ F_{2}---> SF_{6}C

_{4}H_{10}+ O_{2}---> CO_{2}+ H_{2}OS

_{8}+ O_{3}---> SO_{2}

**Solutions:**

^{1}⁄_{8}S_{8}+ 3F_{2}---> SF_{6}C

_{4}H_{10}+^{13}⁄_{2}O_{2}---> 4CO_{2}+ 5H_{2}O

^{1}⁄_{8}S_{8}+^{2}⁄_{3}O_{3}---> SO_{2}

Alternate answer for the third equation:

^{3}⁄_{8}S_{8}+ 2O_{3}---> 3SO_{2}

Another possibility for the third equation:

^{1}⁄_{4}S_{8}+^{4}⁄_{3}O_{3}---> 2SO_{2}

Generally speaking, fractions are mostly used with diatomics (with O_{2} is the most common). In addition, as you delve deeper into balancing, you may even see something like ^{1}⁄_{2}H_{2}O used in a balancing step, but you will never see it as the final answer.

**Example #14a:** H_{2} + O_{3} ---> H_{2}O

However, you must balance it with one restriction: the coefficient in front of the water must be a one.

**Solution:**

1) No fraction to balance the hydrogen:

H_{2}+ O_{3}---> H_{2}OTwo H on the left, two H on the right.

2) Fraction to balance the oxygen:

H_{2}+^{1}⁄_{3}O_{3}---> H_{2}O

^{1}⁄_{3}O_{3}means one O on the left side and there's one O on the right.

Balancing with a ^{1}⁄_{3} as the fractional coefficient is unusual, but you do see it every now and then. Especially, if a teacher is trying to trip you up. Say by using ^{1}⁄_{2} all the time, but then suddenly have a ^{1}⁄_{3} or a ^{5}⁄_{2} on the test. Teachers are sneaky! (Well, at least we think we are!)

**Example #14b:** Na + NH_{3} ---> NaNH_{2} + H_{2} (balance with a fraction)

**Solution:**

Na + NH_{3}---> NaNH_{2}+^{1}⁄_{2}H_{2}And then multiply through by 2.

Notice how the use of the fraction __reduces__ the number of hydrogens on the right-hand side. Usually, you increase the number of atoms on one side to get equality with the other side. However, there ae times where a reduction in an amount can be effectively used. This is one of them.

**Example #15a:** P_{4} + O_{2} ---> P_{2}O_{5}

However, balance it with one restriction: the coefficient in front of the P_{2}O_{5} must be a one.

**Solution:**

1) Balance the P:

^{1}⁄_{2}P_{4}+ O_{2}---> P_{2}O_{5}

2) Balance the O:

^{1}⁄_{2}P_{4}+^{5}⁄_{2}O_{2}---> P_{2}O_{5}

Here's another one to balance:

P_{4}+ H_{2}---> PH_{3}

Keeping the PH_{3} coefficient at one, we have this for an answer:

^{1}⁄_{4}P_{4}+^{3}⁄_{2}H_{2}---> PH_{3}

**Example #15b:** N_{2} + O_{2} ---> N_{2}O_{5}

**Solution #1:**

1) Look at only the oxygen since the nitrogen is already balanced. See that 10 is the least-common multiple between 2 and 5:

N_{2}+ 5O_{2}---> 2N_{2}O_{5}

2) Fix the nitrogen:

2N_{2}+ 5O_{2}---> 2N_{2}O_{5}Done.

**Solution #2:**

1) Use a fraction to balance the oxygen:

N_{2}+^{5}⁄_{2}O_{2}---> N_{2}O_{5}

2) Multiply through by 2 to clear the fraction:

2N_{2}+ 5O_{2}---> 2N_{2}O_{5}Done.

**Example #16:** KFe_{3}AlSi_{3}O_{10}(OH)_{2} + Cu + O_{2} + H_{2}S ---> KAlSi_{3}O_{8} + CuFeS_{2} + H_{2}O

**Solution:**

Looks pretty tough, doesn't it? Before balancing, examine the equation and note:

a) K, Al, and Si are already balanced.

b) Balancing Fe is easy.

c) The H is in only one place on the right.

d) O is everywhere. Leave it to the last.

1) With that in mind, balance the Fe, then the Cu:

KFe_{3}AlSi_{3}O_{10}(OH)_{2}+ 3Cu + O_{2}+ H_{2}S ---> KAlSi_{3}O_{8}+ 3CuFeS_{2}+ H_{2}O

2) Now, the S:

KFe_{3}AlSi_{3}O_{10}(OH)_{2}+ 3Cu + O_{2}+ 6H_{2}S ---> KAlSi_{3}O_{8}+ 3CuFeS_{2}+ H_{2}O

3) The K, Al, and Si are already balanced, leaving only H and O to be dealt with. Of the two, H is the easiest one, with a total of 14 on the left. So, a 7 on the right balances the H:

KFe_{3}AlSi_{3}O_{10}(OH)_{2}+ 3Cu + O_{2}+ 6H_{2}S ---> KAlSi_{3}O_{8}+ 3CuFeS_{2}+ 7H_{2}O

4) Now, count oxygens on the right to get 15. Count on the left to get 14. We need one more oxygen on the left:

KFe_{3}AlSi_{3}O_{10}(OH)_{2}+ 3Cu +^{3}⁄_{2}O_{2}+ 6H_{2}S ---> KAlSi_{3}O_{8}+ 3CuFeS_{2}+ 7H_{2}ONote that I used a

^{3}⁄_{2}. That is because I needed one more oxygen atom in addition to what was already there. What was already there was one O_{2}molecule (think of it as^{2}⁄_{2}O_{2}). An increase from^{2}⁄_{2}to^{3}⁄_{2}is one more atom.

5) Multiply through by two to get the final answer:

2KFe_{3}AlSi_{3}O_{10}(OH)_{2}+ 6Cu + 3O_{2}+ 12H_{2}S ---> 2KAlSi_{3}O_{8}+ 6CuFeS_{2}+ 14H_{2}O

**Example #17:** SiO_{2} + CaC_{2} ---> Si + CaO + CO_{2}

**Solution:**

1) Balance the C:

SiO_{2}+ CaC_{2}---> Si + CaO + 2CO_{2}

2) The only element left to balance is the oxygen. I'll use a fraction to balance it:

^{5}⁄_{2}SiO_{2}+ CaC_{2}---> Si + CaO + 2CO_{2}

3) Balance the Si:

^{5}⁄_{2}SiO_{2}+ CaC_{2}--->^{5}⁄_{2}Si + CaO + 2CO_{2}

4) Multiply through to clear the fraction:

5SiO_{2}+ 2CaC_{2}---> 5Si + 2CaO + 4CO_{2}

You may have protested at the ^{5}⁄_{2} used with the Si. You might have said "But you can't have fractional atoms" or "But you can't have ^{5}⁄_{2} of an atom."

Those statements are true, but notice that the ^{5}⁄_{2} is simply being used as a step along the way to the final answer. The ^{5}⁄_{2} is not being represented as the final answer.

Also, there is an additional understanding of chemical equations that you might not yet be aware of. Chemical equations can be understood on the basis of a chemical measurement called the 'mole.' It is perfectly proper to say that I have ^{5}⁄_{2} of a mole of Si.

If you're not sure exactly what a mole is yet, that's OK. Give it a bit more time and study.

**Example #18:** NH_{3}(g) + O_{2}(g) ---> NO_{2}(g) + H_{2}O(g)

**Solution:**

1) The nitrogens are already balanced, so leave them alone. The oxygens are in three of the four compounds whereas hydrogen is in only one reactant and one product. And, it's not balanced. Balance the hydrogen:

2NH_{3}(g) + O_{2}(g) ---> NO_{2}(g) + 3H_{2}O(g)Remember, 6 is the least common multiple between 2 and 3.

2) Balancing the hydrogens has put the nitrogen out of balance. Balance the nitrogen:

2NH_{3}(g) + O_{2}(g) ---> 2NO_{2}(g) + 3H_{2}O(g)

3) We are now ready to balance the oxygen:

2NH_{3}(g) +^{7}⁄_{2}O_{2}(g) ---> 2NO_{2}(g) + 3H_{2}O(g)

4) Multiply through by 2 to clear the fraction:

4NH_{3}(g) + 7O_{2}(g) ---> 4NO_{2}(g) + 6H_{2}O(g)

5) Notice that the usual way to balance is to leave oxygen/hydrogen to the end. That rule is "violated" from time to time and this equation is a good example of that.

**Example #19:** MgNH_{4}PO_{4} ---> Mg_{2}P_{2}O_{7} + NH_{3} + H_{2}O

**Solution:**

1) Balance the magnesium:

2MgNH_{4}PO_{4}---> Mg_{2}P_{2}O_{7}+ NH_{3}+ H_{2}OThis also balances the phosphorous.

2) Balance the nitrogen:

2MgNH_{4}PO_{4}---> Mg_{2}P_{2}O_{7}+ 2NH_{3}+ H_{2}O

3) Examine the situation with hydrogen and oxygen and discover they are both balanced with 8 of each. The equation is now balanced.

**Example #20:** Balance:

PCl_{5}+ H_{2}O ---> H_{3}PO_{4}+ HCl

using the LCM method.

**Solution:**

When I first saw this question on Yahoo Answers, I thought "OK, some least common multiple action. I know how to do that:

PCl_{5}+ 3H_{2}O ---> 2H_{3}PO_{4}+ HCl

As I looked at this, planning my next step, it suddenly dawned on me: the question writer (who is not the ChemTeam) is trying to mess with his/her students' head by specifying the use of LCM.

There really isn't a good path to balancing this equation via LCM. After all, the LCM step above doesn't even balance the H. And, if you tried to balance the H, you'd mess it up even more. And the oxygen is totally messed up.

You should use a non-LCM method.

Balance by way of balancing the oxygen first. Like this:

PCl_{5}+ 4H_{2}O ---> H_{3}PO_{4}+ HCl (oxygen balanced)PCl

_{5}+ 4H_{2}O ---> H_{3}PO_{4}+ 5HCl (hydrogen balanced, chlorine also)

Balance by way of balancing the chlorine first. Like this:

PCl_{5}+ H_{2}O ---> H_{3}PO_{4}+ 5HCl (chlorine balanced)PCl

_{5}+ 4H_{2}O ---> H_{3}PO_{4}+ 5HCl (hydrogen balanced, oxygen also)

The following equation is another that doesn't fit well with the LCM idea:

PCl_{3}+ H_{2}O ---> H_{3}PO_{3}+ HCl

**Bonus Example #1:** Balance:

K_{4}Fe(CN)_{6}+ H_{2}SO_{4}+ H_{2}O ---> K_{2}SO_{4}+ FeSO_{4}+ (NH_{4})_{2}SO_{4}+ CO

**Solution:**

Note: it looks like it might be a complicated redox equation involving three half-reactions (based on the Fe, the C and the N). However, when you examine the oxidation state of each element, you find that nothing changes. (By the way, if you have no clue what an oxidation state or a half-reaction is, don't worry about it. You'll learn them soon enough!)

1) K, Fe, N and C all go from one reactant to one product. Balance them all:

K_{4}Fe(CN)_{6}+ H_{2}SO_{4}+ H_{2}O ---> 2K_{2}SO_{4}+ FeSO_{4}+ 3(NH_{4})_{2}SO_{4}+ 6CO

2) Sulfate is in only one place as a reactant. Add up the sulfates on the right-hand side and balance:

K_{4}Fe(CN)_{6}+ 6H_{2}SO_{4}+ H_{2}O ---> 2K_{2}SO_{4}+ FeSO_{4}+ 3(NH_{4})_{2}SO_{4}+ 6CO

3) The only elements left are the H in the ammonium sulfate and the O in the carbon monoxide. Count them up to get 24 H and 6 O. 12 H are accounted for in the sulfuric acid, so balance the rest (12 H and 6 O) using the water on the left:

K_{4}Fe(CN)_{6}+ 6H_{2}SO_{4}+ 6H_{2}O ---> 2K_{2}SO_{4}+ FeSO_{4}+ 3(NH_{4})_{2}SO_{4}+ 6CO

**Bonus Example #2:** C_{3}H_{7}S(ℓ) + O_{2}(g) ---> CO_{2}(g) + H_{2}O(ℓ) + SO_{2}(g)

**Solution:**

1) Balance the carbon:

C_{3}H_{7}S(ℓ) + O_{2}(g) ---> 3CO_{2}(g) + H_{2}O(ℓ) + SO_{2}(g)

2) Balance the hydrogen with a fraction:

C_{3}H_{7}S(ℓ) + O_{2}(g) ---> 3CO_{2}(g) +^{7}⁄_{2}H_{2}O(ℓ) + SO_{2}(g)

3) Clear the fraction:

2C_{3}H_{7}S(ℓ) + 2O_{2}(g) ---> 6CO_{2}(g) + 7H_{2}O(ℓ) + 2SO_{2}(g)

4) Count oxygens on right-hand side to find 23. Use a fraction to balance:

2C_{3}H_{7}S(ℓ) +^{23}⁄_{2}O_{2}(g) ---> 6CO_{2}(g) + 7H_{2}O(ℓ) + 2SO_{2}(g)

5) Clear the fraction:

4C_{3}H_{7}S(ℓ) + 23O_{2}(g) ---> 12CO_{2}(g) + 14H_{2}O(ℓ) + 4SO_{2}(g)

Probs 1-10 | Probs 11-25 | Probs 26-45 | Probs 46-65 |

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