Balancing Equations This tutorial will just take you through several balancing equation problems, starting with some pretty easy ones, and then getting progressively more and more difficult.

Balancing Equations 1. __ H 2 + __ F 2  __ HF Starting with the H’s, we have two on the left. and only one on the right. So we need to put a “2” on the HF. That gives us two H’s on both sides. Now the F’s: We have two on the left, And we have two on the right, because the “2” we put in front of the HF pertains to the entire HF molecule – two H’s and 2 F’s. So everything is balanced. 2 H F H F

F Balancing Equations Starting with the N’s, we have two on the left. and only one on the right. So we need to put a “2” on the NF 3. That gives us two N’s on both sides. Now the F’s: We have two on the left, And how many F’s on the right? So we need to put a 3 on the F 2. That gives us six F’s on both sides. Now everything is balanced. 2 N F 3 N F F F F three groups of 2= two groups of 3 6 If you were thinking 3, you forgot about the 2 we put in front of the NF 3. 2 x 3 = __ N 2 + __ F 2  __ NF 3

Balancing Equations 3. __ Na 2 S + __ FeBr 3  __ NaBr + __ Fe 2 S 3 Starting with the Na’s, we have two on the left. and only one on the right. So we need to put a “2” on the NaBr. That gives us two Na’s on both sides. Now the S’s: We have one on the left, And three on the right? So we need to put a 3 on the Na 2 S. That gives us three S’s on both sides, But it messes up the Na’s: now we have six Na’s on the left. So we need to change the 2 NaBr to 6 NaBr. Since you often have to change numbers you’ve written down, it’s always a good idea to use a pencil or erasable pen when balancing equations! If you were thinking 3, you forgot about the 2 we put in front of the NF 3. 2 x 3 = 6. 6 three groups of 2= six groups of 1

Balancing Equations 3. __ Na 2 S + __ FeBr 3  __ NaBr + __ Fe 2 S 3 So now the Na’s are balanced again with six on each side. Now the Fe’s: we have one Fe on the left and two Fe’s on the right. So what should we do? Now we have 2 Fe’s on both sides. Finally the Br’s! How many Br’s do we have on the left? And how many Br’s are on the right? So it seems everything is balanced, but with all the back and forth and changes made, it’s best to check our answer. Go ahead and “take inventory” on both sides. You should have counted six Na’s, three S’s, two Fe’s and six Br’s on each side… so it is balanced! 3 6 If you were thinking 3, you forgot about the 2 we put in front of the NF 3. 2 x 3 = 6. 6 Put a 2 on the FeBr 3. 2 If you were thinking just 3, again you were forgetting about the 2 we put on the FeBr 3. Remember always to look both in front of and behind the atoms you are balancing. Also 6 Sorry, I’m not going to show you that little rearrangement of the atoms this time… Hopefully you can see that the equation is balanced by just looking at the formulas and coefficients above!

9. __ SnF 4 + __ S 8  __ SnS 2 + __ F 2 Balancing Equations 4. __ H 2 + __ O 2  __ H 2 O So now let’s practice a few relatively easy equation balancing problems and see how you do. Try these on your own, one at a time, and then advance the slide to see if you are right. If you get it wrong, try to see where you made your mistake. 5. __ K + __ N 2  __ K 3 N 6. __ Fe + __ O 2  __ Fe 2 O 3 7. __ AlBr 3 + __ I 2  __ Al I 3 + __ Br 2 8. __ PF 3  __ P 4 + __ F

Balancing Equations 10. __ K 3 BO 3 + __ Sn(C 2 H 3 O 2 ) 4  __ KC 2 H 3 O 2 + __ Sn 3 (BO 3 ) 4 Right away you are probably looking at this one and thinking that it’s going to be really difficult. But there is a “trick of the trade” we will use to make it much easier. Do you notice that there are polyatomic ions that appear on both sides? Rather than breaking them up into their atoms, we are going to keep them together and balance them as bundles. For example, rather than breaking up the first one into B’s and O’s, we will simply ask: “How many BO 3 bundles are on the left side, and how many are on the right side?” Answers: one on the left and four on the right. (Keep in mind, the “3” is part of the BO 3 bundle.)

Balancing Equations 10. __ K 3 BO 3 + __ Sn(C 2 H 3 O 2 ) 4  __ KC 2 H 3 O 2 + __ Sn 3 (BO 3 ) 4 So, let’s start with the K’s. How would you balance them? Hopefully you saw there were three on the left and only one on the right and so you put a 3 on the KC 2 H 3 O 2. Now, as we said before, we have one BO 3 on the left and four on the right. What should we do to balance the BO 3 ’s? Putting a 4 on the K 3 BO 3 makes the BO 3 ‘s balanced but… Now we have twelve K’s on the left. What should we do? Were you thinking of changing the 3 on KC 2 H 3 O 2 to a 12? Good, so now how do we balance the Sn’s? It will require a 3 on the Sn(C 2 H 3 O 2 ) 2 to balance them. Finally, how many C 2 H 3 O 2 bundles do we have on the left? If you were thinking four, check again. twelve C 2 H 3 O 2 ’s on the right as well, the equation’s balanced It’s twelve. And with

Balancing Equations 11. __ (NH 4 ) 2 SO 4 + __ Pb(NO 3 ) 4  __ Pb(SO 4 ) 2 + __ NH 4 NO 3 So now let’s practice a few equation balancing problems involving these polyatomic ions. Remember trick of the trade #1: Keep polyatomic ions together and balance them as bundles. 12. __ Al 2 (CO 3 ) 3 + __ HClO 4  __ Al(ClO 4 ) 3 + __ H 2 CO __ Ca(NO 3 ) 2 + __ K 3 PO 4  __ Ca 3 (PO 4 ) 2 + __ KNO __ (NH 4 ) 3 BO 3 + __ Sn(BrO 3 ) 4  __ Sn 3 (PO 4 ) 4 + __ NH 4 BrO

Balancing Equations 15. __ Na + __ H 2 O  __ NaOH + __ H 2 Sometimes when H 2 O is in a simple equation like the one above, it may prove difficult to balance. Trick of the trade #2: If you find yourself having trouble balancing an equation that has H 2 O in it, try treating the H 2 O like H(OH) instead, and then balance the H and the OH completely separately: you may find it easier to balance.

Balancing Equations 15. __ Na + __ H 2 O  __ NaOH + __ H 2 So, in the equation above, we will try this: We can see we have one Na on each side, so they’re good. How many H’s do we have on the left? If you were thinking two, then you were forgetting, we are balancing the H’s and the OH’s separately. And how many H’s do we have on the right? Remember: don’t count the H in Na(OH). So let’s put a 2 on the H(OH). Now we have two OH’s on the left & only one on the right. So we must put a 2 on the NaOH. So are we all done? Now the Na’s need to be balanced again! H(OH) Just one. Two. 2 2 Not quite. 2 Now we’re done!

Balancing Equations 16. __ H 2 O + __ Ba  __ H 2 + __ Ba(OH) 2 So now let’s practice a few problems like that last one. Remember trick of the trade #2: Sometimes it’s useful to treat H 2 O as H(OH) and balance the H and the OH separately. 17. __ Cr + __ H 2 O  __ Cr(OH) 3 + __ H __ H 2 SO 4 + __ Fe(OH) 3  __ H 2 O + __ Fe 2 (SO 4 )

Balancing Equations 19. __ O 2 + __ C 3 H 8  __ CO 2 + __ H 2 O This problem may look pretty easy, but there is one slight speed bump you need to be aware of. Whereas in all the problems we have tried so far, each element (or polyatomic ion bundle) has appeared in just two places – one on each side of the equation– here we have an element that occurs in three places: O Trick of the trade #3: When an element appears in more than two places in an equation, always save it for last.

Balancing Equations 19. __ O 2 + __ C 3 H 8  __ CO 2 + __ H 2 O So rather than begin with the O’s, we will skip them and start with the C’s. How would you balance the C’s? Hopefully you saw that there were three C’s on the left and only one on the right, so you put a 3 on the CO 2. Now let’s balance the H’s… Any ideas? If you saw that there were eight H’s on the left and only two on the right, you probably knew you needed a 4 on the H 2 O. Now the O’s. You can see that there are two on the left side, but how many O’s are there all together on the right? Your answer should have been ten. So with ten O’s on the right and only two on the left, you should know what to do: 34 5 sixplus four

Balancing Equations 20. __ C 5 H 8 + __ O 2  __ CO 2 + __ H 2 O So now let’s practice a few equation balancing problems involving elements appearing more than twice. Remember trick of the trade #3: If an element appears more than twice in the equation, save balancing that element for last. 21. __ C 3 H 8 + __ F 2  __ CF 4 + __ HF 22. __ N 3 H 5 + __ O 2  __ N 2 O + __ H 2 O 23. __ C 9 H 18 O + __ O 2  __ CO 2 + __ H 2 O

Balancing Equations 24. __ C 4 H 10 + __ O 2  __ CO 2 + __ H 2 O This problem looks pretty much like the last ones you just did, with one slight difference. The easiest way to balance it requires a fraction. Yet balanced equations are most often written with whole numbers only. Trick of the trade #4: If you need to use a fraction to balance an equation, do it, but then multiply everything by the denominator (usually 2) to get rid of the fraction.

Balancing Equations 24. __ C 4 H 10 + __ O 2  __ CO 2 + __ H 2 O First we will balance the C’s. Next let’s balance the H’s. There are two O’s on the left; how many O’s on the right? Hopefully you counted thirteen. And the only place we can put a number that won’t throw everything else off is in front of the O 2. But how do you change a 2 into a 13? Answer: by multiplying it by 6.5 …or 6 1 / 2 …or 13 / 2. But because we can’t leave fractions in a balanced equation, we will now multiply every coefficient by the fraction’s denominator: 2. And now the equation is balanced / 2 eight plus five

Balancing Equations 25. __ C 2 H 6 + __ O 2  __ CO 2 + __ H 2 O So now let’s practice a few final balancing problems involving fractions. Remember trick of the trade #4: Use fractions if you need to, but then multiply through to get rid of them. 26. __ C 5 H 10 + __ O 2  __ CO 2 + __ H 2 O 27. __ N 4 H 6 + __ O 2  __ NO 2 + __ H 2 O 28. __ Fe 3 P 2 + __ P 4  __ FeP __ C 3 H 8 O + __ O 2  __ CO 2 + __ H 2 O 30. __ C 6 H 14 O 2 + __ Fe 2 O 3  __ CO 2 + __ H 2 O + __ Fe These last three are pretty tricky, but try your best!