Chemistry HW/CW [1/3/2011] – Balancing Chemical Equations

For quick finding, here are some Find (Ctrl F) codes on this page only [cool concept but hard to execute every single time; it's like typing up a FAQ]:

• (W001) – Beginning of Work Section.
  
• (W002) – First mention of Reactants and Products with proceeding example
  
• (W003) – First Step of the First Problem
  
• (W004) – Second Step of the First Problem; first attempt at balancing
  
• (W005) – Third and Final Step of the First Problem; second and last attempt at balancing.
  
• (W006) – End Result
  
• (CH001) – Beginning of Challenge Problem
  
• (CH002) – Counting Challenge Atoms
  
• (CH003) – First Step of Challenge
  
• (CH004) – Second Step of Challenge
  
• (CH005) – Third Step of Challenge
  
• (CH006) – Fourth Step of Challenge
  
• (CH007) – Fifth Step of Challenge
  

Introduction: It's been quite a while since I've posted anything here, so here's my first post in a while! The Chemistry homework/classwork for January 3rd, 2012.

Work (W001): I'll only do Part C because B and A are pretty self explanatory.
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What you're trying to do in the first place is place subscripts to "complete" the chemical equation, as you never get more/less output than you give input (it's a physics law, can't remember which one). So, here is the first problem:

V (Reactants side)..............................V (Products side) (W002)

1) P O₂ --> P₄O₁₀


Okay, so the job here is to make that thing balanced.

Step One: Count the number of atoms you have. Put a bar or something to denote reactants/products count. (W003)

P (phosphorus): 1 | 4
O (oxygen): 2 | 10
]P + O₂ --> P₄O_10]
That part is all done and over with. Now, does 1 equal 4? No, it doesn't. Does 2 equal 10? No, it doesn't. So, we'll have to do some balancing, aka, placing coefficients. On the REACTANTS side (always the left), we'll put a FOUR [4] next to P. That now makes the count as (W004)

P (phosphorus) 4 | 4
O (oxygen): 2 | 10

s]P + O₂ --> P₄O₁₀]

So now, we have to balance Oxygen. Let's place a FIVE (5) next to O₂. That now makes the count as (W005)

P (phosphorus): 4 | 4
O (oxygen): 10 | 10

s]P + 5O₂ --> P₄O₁₀]

The end result will look like this (W006):

4P + 5O₂ ----> P₄O₁₀

FINISH, SUCCESS. Sometimes, it gets complicated and you'll have to dabble with the PRODUCTS side in order to balance the equation. One such complicated problem? The challenge, which is on the back of the paper. In fact, let's solve that one.

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Challenge (CH001)

The formidable challenge question. The one that makes everyone go "WTFBOOM". Well, that's completely unnecessary because a guide could be posted! Here we go!!

The equation looks a little something like this:

C₂H₆ + O₂ ---> CO₂ + H₂O

So, to first step once again, count up the number of atoms (CH002):

C: 2 | 1
H: 6 | 2
O: 2 | 3

What oh what shall we do first? Oh, I know! Let's add a 2 coefficient to C₂H₆. Yes, yes, you're probably saying "But that solves nothing at all!". You'll thank me later, trust me! So now the count is and the formula looks like (CH003):

C: 4 | 1
H: 12| 2
O: 2 | 3

2C₂H₆ + O₂ ---> CO₂ + H₂O

Okay, let's do a bit more manipulation. Yes, yes, you're going to say "WHY?!" but you'll see the result in due time! Let's add a SEVEN [7] coefficient to O₂. So now, the count is and the formula looks like (CH004):

C: 4 | 1
H: 12 | 2
O: 14 | 3

2C₂H₆ + 7O₂ ---> CO₂ + H₂O

Things are getting confusing, yes, but watch this: add a FOUR [4] coefficient to CO₂, thus bringing the count to and formula to appear as (CH005):

C: 4 | 4
H:12 | 2
O: 14 | 9

2C₂H₆ + 7O₂ ---> 4CO₂ + H₂O

Now we're going to add a SIX [6] coefficient to H₂O. Result (CH006):

C: 4 |4
H:12 | 12
O: 14 | 14

2C₂H₆ + 7O₂ ---> 4CO₂ + 7H₂O

FINISH, SUCCESS! End Result (CH007):

2C₂H₆ + 7O₂ ---> 4CO₂ + 7H₂O

No need to thank me at all! But you can always acknowledge that you looked at this site by registering!

will post all challenge problems, introduction and solutions inshAllah soon.
If I didn't, something else took my attention.