Matrix Algebra

 On using a combination of tricks to solve matrix algbra.  One solving this matrix involves the use
of multiplying by a fraction and two for one of the new rows you have
to multiply row 1 by something and row 2 by something THEN add them
together.  For your math pleasure included below in the row operations
I did to solve the matrix.  Follow along and verify that I did my math
right.

The original equations:

X + Y -Z = 6
3X -2Y +Z = -5
X + 3Y -2Z =14

Which produces the augemented matrix

[1  1 -1 :  6]
[3 -2  1 : -5]
[1  3 -2 :14]

I want my matrix to look like

[1 0 0 : a]
[0 1 0 : b]
[0 0 1 : c]

So the first thing I need a 1 in the upper left corner.  Check.  That was easy.
I now need 0s.  How do I get zeros?  If I used the operations:

R2 = -3r1 +r2    &
R3 = r1 + (-1)r3

I would end up with the matrix
[1  1 -1 :    6]
[0 -5  4 : -23]
[0 -2  1 :   -8]

Remember to double check what I am doing.  It does you NO good if I
understand how to do this.

This next part I can see where you would be confused where to go next.
 I need a zero on the bottom.  How do I get it there?  What is a
common denominator between 5 and 2?  Would 10 work?  How do I make
each fisrt number 10?

What if I times row 2 by -2 and row 3 by +5  Then I could add -10 to
+10 and would that = 0?

So try the row operation:

R3 = -2r2 + 5r3

If you do, you should get the matrix
[1  1 -1 :    6]
[0 -5  4 : -23]
[0  0  3 :    6]

Email me if you have any questions about that last step and how I got
it. If you get it we'll continue on.

At this point I can either get ones along the diagonal or go for
zeros.  Also at this point if the teacher didn't ask for reduced row
echelon (r.r.e) form I am actually done.  Because I know that

3Z = 6 solve for Z

-5Y + 4Z = -23 (I know Z from above so plug in and solve for Y).

& X + Y -Z (Again from above I now know Y and Z so plug in and get X).

But assuming your teacher was a jerk and actually wanted r.r.e form
we'll continue our dreary task.

Because it is super easy I am going to make the diagonal 1s.

let
R2 = (-1/5)r2  &
R3 = (-1/3)r3

If we do this our new matrix is

[1 1   -1    :       6]
[0 1 (-4/5) : (23/5)]
[0 0    1    :      -2]

Now let R3 = r2+ (4/5)r3.  I leave it to you to figure out why.  It is
also left up to you to see if there was another way of proceeding if I
hadn't changed the diagonals to ones.  I'll give you a hint.  It
involves +/- 12 and its common factors.

[ 1 1 -1 : 6]
[ 0 1  0 : 3]
[ 0 0  1 : 2]

I purposely dropped a negative sign in the above matrix, see if you
can find it.  Don't go on until you find it.  I want to make sure you
are paying attention and not just mindlessly reading my email.

Found it yet?  Good!

For my next trick I will let R1 = r1 + r3.  Why?  To get a zero where I want it.

[1 1 0 :  4]
[0 1 0 :  3]
[0 0 1 : -2]

I am almost there.  I only have one more set of row operations.  I am
going to let you figure out what they are.  But if done right you
should get the matrix

[1 0 0 :  1]
[0 1 0 :  3]
[0 0 1 : -2]

Did you get the same thing I did?  Do you understand what I did and
more importantly, why I did it?  Is there anything that didn't make
sense that I need to go over one more time.  Speak now or forever hold
your peace, well at least until next time.  I hope that helped.  If
you need anything else you know how to reach me.

Questions regarding O-Chem

1. Hydrogen is the antithesis to double and triple bonds, i.e. hydrogen is kryptonite to the double and triple bonds superman.  Since there is only 1 mole of H2 not all of the triple bond can break only one of them.  Therefore the major product should be a.  A chain with seven carbons with a double bond in the middle i.e. choice a.  C. lost a carbon somewhere.  i would go with a.  If there were lots and lots of hydrogen d is possible but unlikely given only 1 mole.

 

2. Aromatic means that there is an even number alternating single and double bonds equal to 4n+2 where n=0,1,2,3,...

http://en.wikipedia.org/wiki/Toluene  All of these compounds are real but they go by aliases (you'll probably have to memorize these.) Counting carbons toluene has only 7 so it won't work.

http://en.wikipedia.org/wiki/Xylene  the "o" is NOT oxygen it actually refers to the placement of the methyl groups.  If two groups are next to each other it is called ortho-whatever or o-whatever.  If there is a carbon between the two side groups then its called para or p-whatever.  And if there are two carbons between side groups then its called meta or m-whatever.  Looking at the picture this clearly has 8 carbons.  C. Will NOT work because dichloro added two chlorine atoms chlorine is NOT carbon therefore dichlorobenzene has only 6 carbons.

D. phenyl means it has 6 carbons butan means means it has 4 carbons for a grand total of 10 carbons i.e. too many.

E. Ethyl means two carbons.  Benzene means 6 with three double bonds i.e. 8 carbons so this one is good.

F. Has 12 carbons.

 

Thus for number 2 I would pick B and E. (Using Wiki is ok as long as its not during a test.)

 

3. How long is the longest chain.  6.  Does it have any double bonds?  Yes.  How many?  Three.  Does this structure have a common name?  Yes.  Benzene.  Thus whatever it is it ends in something benzene.  Which is more important, boring hydro carbon chains or exotic halides.  Correct Halides.  Is either halide more important? No.  So which one comes first alphabetically, chloro or bromo?  Bromo good.  So number starting at bromine.  Now I can either number clock wise or anti clockwise.  Which choice will give us smaller numbers on more pieces.  Good so let's number them clockwise.  How long is that carbon chain?  Good 3.  What do they call a three carbon chain?  Prop.  Good now is it the main chain or a side chain? Is it bigger than six or smaller?  Its smaller.  Good so it is a side chain.  What do side chains end with?  "yl"  Good.  Thus we start numbering and naming.  Put the halides first then the alkane group. And voila! 1-bromo-3-chloro-2-propylbenzene.

 

4.  I think they all will react, but I would check the chapter to make certain.

 

5.  http://en.wikipedia.org/wiki/File:Benzene_Representations.svg

Hybridization is a lumping together of two electron orbitals.  Think of them as the race tracks electrons follow.  S orbitals look like spheres.  P orbitals look like figure 8s or dumbells.  an sp orbital is bulbous on one end and skinky on the other as though a p orbital swallowed the spherical s orbital.  Have you ever had room mates.  There is what is distinctly your stuff.  Then there is the stuff that is distinctly their stuff, but then there's some of the stuff that you have NO idea whose it is.  When dealing with electrons its the same way.  When the two orbitals merge they call them hybridized orbitals because its a combination of s stuff and p stuff and has kind of a unique shape much like a common room in a shared apartment.  Anyway you have only a few choices for hybridized orbitals.  sp3 has four lobes so something like methane

 

   H

    |

H-C-H

    |

   H

 

One central atom and four thingamajigs coming off of it be they hydrogens, lone pair electrons, bromines, big long hair carbon chain or what not there are four things, its doesn't matter what they are all coming off of the central atom.  we call this sp3 hyridization and it is really easy to remember because you can think of it as s1p3 1+3=4 so an sp3 hybrid should have four lobes.  Now its your turn.  How many lobes will a sp2 hybrid have?  If you are paying attention it should have one less than an sp3 so sp2 have a total of not four but 3 lobes.  This is easily double checked by thinking of sp2 as s1p2 where 1+2= you guessed it--3.  So an sp2 hybrid should have three lobes not four.  Carbons in double bonds often have this arrangement.  Consider the following:

 

    H        H

      \      /

#1   C=C  # 2 

      /       \

   Br        CH2CH2CH2CH3

 

 

Carbon #1 has a C, a H, and a Br bonded to it forming three lobes.  Yes one of the carbons is double bonded so there are really two attachments but if you were to circle all of the stuff attached to carbon #1 you would draw three circles.  Thus the hybridization must add up to three lobes sp2.  Similarly Carbon#2 also has three things attached to it.  One double bonded carbon forming one lobe.  One hydrogen atom forming the second lobe.  And one LONG chain of crap I don't care about forming a third lobe.  (If the question had asked about some of the other carbons I might care and carefully draw out the carbons and hydrogens and I would see they all have four things bonded to them making them sp3 like we learned about earlier.)  Notice that when dealing with hybridizations its not the total number of bonds because in carbon that is always 4, but rather the important part is how many different directions are the bonds facing.

 

Last example.

 

Consider the molecule aceylene aka ethyne

 

H-C*C-H

 

Where * is a triple bond between the carbons.  In this case the bonds go only two different directions I would thus say it has two lobes.  Therefore whatever hybridization is going on must add up to 2.  Well, if sp3 adds up to 4.  And sp2 adds up to 3.  Then sp must add up to two.  Sure enough  we think of it as s1p1 1+1=2 Thus if a molecule has only two lobes, i.e. a triple bond it must be sp hybridized.

 

Brief recap.  sp3=4 lobes i.e. all single bonds.  sp2=3 lobes i.e. a double bond.  sp=2 lobes and is usually a triple bond.  NOTE:  This only applies to the carbons.  Hydrogens only have 1 thing attached to them so they will ALWAYS be unhybridized s orbitals.  Hydrogen is like your underwear.  As much stuff as you share and hybridize with your room mates your underwear is probably not among those things.  Hydrogen stays unhybridized carbon's a bit of a swinger with its electrons.

 

Now let's go back to the question. 

 

Benzene is often depicted as something like this:

 

 /  \

| O |

 \  /

 

but you should think of it as something like

                                                                                      

                                                                   

       H 

        |

       C

      /  \\     

H-C     C-H

   ||       |

   C     C-H

  /   \   //

H      C

         |

        H

or    

       H

        |    

       C

      //  \                                                                               

H-C     C-H                                                                                                             

   |       ||                                                              

   C     C-H                                                                           

  /   \\   /                                                         

H      C                                                             

         |

        H    

 

With the double bond bouncing everywhere faster than is possible to measure.  Since each of the carbons has three lobes we could call it sp2 hybridized, but there is a better description.  Delocalized electrons.  I would call it e.  There are no triples, not sp orbitals.  The hydrogens are on the outside not on the ring, there are no chlorines or other halogens thus e is probably the best choice.

Significant Figures

Dear Tutees,

For your viewing and studying pleasure, today's post is an excerpt from my Master Tutor Project.  Enjoy.

This is an idea I got from my high school chemistry teacher for remembering placement of the decimal in significant figures.    It is called the Atlantic and Pacific rule.

Most people will recognize that the “P” acific Ocean is on the left, while the “A” tlantic Ocean is on the right.  If a decimal is “P” resent count all non-zeros starting at the left.  If a decimal is “A” bsent count all non-zeros starting at the right.  Any bodies of water between the Atlantic and Pacific are ignored, just as any zeros in the middle of a number are ignored.  (Image Source: http://www.theodora.com/maps/new3/north_america_color.gif).   Examples:

503,000  Note that a decimal point is absent.  Therefore, start on the right count all non-zeros e.g. 503 thus 3 significant figures.  But, 503,000.00 has a decimal point present, so start at the Pacific or the left side or 8 significant figures. But 0.001 has only 1 because the first three zeros are non-significant.

About the Ads

A Personal Message From the CEO of John Boom Tutoring:

Dear Loyal Readers,

Previously, on other blogs of mine you may have read,  that I went to great trouble chasing away the ferocious ad beasts.  However, since this is my business page and not my personal page, I do not in the least feel that I have "sold out" by allowing ads here, for that is this page's purpose--to make money.  Hopeful readers fear not, for I still staunchly defend my other blog-homes keeping them safe from whiles of the ad-paparazzi, but while they may, ads may safely rest here.  Be aware though, I do not endorse, nor agree with, nor even necessarily know anything about what is being sold here.  Click at your own discretion.  I will try to keep it family appropriate and minimally annoying.  It is after all, my blog too.

Sincerely,
John Boom-CEO and Owner John Boom Tutoring

P.S Send me your feedback.  Let me know what you think.  If there is something you are struggling with post a comment somewhere and I will put up a reply.  Best of luck!  Hope all is well.  Cheers to you.

All Possible Graphs

The above represent a sampling of all possible variations of the basic graphs you will encounter in College Algebra, including: Sin, Cos, Tan, Log, Exponential, Abs, Linear, Squared, and Cubic graphs.  Feel free to print, use and share as you would like.  However, I do ask that you please leave the attribution tag "JOHNBOOMTUTORING.BLOGSPOT.COM" on any materials of mine you produce.  Thank you and good luck!

Variations on Common Inverse Trig-Integrations

The following are variations  of the common Inverse Trigonometric Integrals starting with ArcTan, moving to ArcSec, and concluding with the ArcSin. 

Solving things like the integrals above is relatively "easy"  Once you pick up the pattern.  They usually require only applications of the Pythagorean identity and some creative algebra.  That is…

A² + B² = C² or in trig speak…

Sinθ² + Cosθ² = 1

Which if we divide both sides by Cosθ² can be rearranged as

Tanθ² + 1 = Secθ²

Thus giving us the three identities we need to solve the above problems.

x² + 1 = Tanθ² + 1 = Secθ²

x² – 1 = Secθ² -1 = Tanθ²

1-x² = 1-  Sinθ² = Cosθ²

Thus if we see x² +1 we want to set x = Tanθ with the associated implications that θ = Tan^-1 x and dx = sec²θ.  We do this because if x = Tanθ we can totally rock the identity we derived that Tanθ² + 1 = Secθ².  Similarly if we see things of the form x² – 1 we will want to let x = Secθ with dx = Secθ Tanθ dθ.  And finally, if you see 1-x² let x = Sinθ and let dx = Cosθ.  The use of trig identities was, after all, the entire reason we made the wacky substations in the first place was so we could get some trig identity action working in our favor.  Making these substitutions leaves us with…

(  ___dx___    = (  __ sec²θ  dθ   = (   dθ = θ + C = ArcTan x +C or Tan^-1x + C

 )     x² + 1             )     sec²θ               )

Sweet, letting x = Tanθ was a pretty good idea, especially when combined with that trig identity we tossed in there canceling the Secθ² !  NOTE:   letting X = Tanθ also works if I have 1+X²     because A+B is the same as B+ A.  

This next one though it looks simple is a bit messier.  But let’s dive in.

(  ___dx___   

  )     x² – 1

This time let X = Secθ.  If we do this I can get this to simplify using Secθ² -1 = Tanθ².  If x = Secθ this means θ = Sec^-1x.  It also means that dx = d(Secθ).  I always remember the derivative of secant with the phrase "dseek tan destroy" playing off of the phrase seek and destroy representing the answer to the derivative of secant  = SecθTanθ dθ.  Making these switches leaves us…

(  ___dx___  = (  __ SecθTanθ dθ ___= (  __ SecθTanθ dθ ___=  (  __ Secθ dθ_ =(__1/Cosθ__dθ  

  )     x² – 1        )        Secθ² -1               )              Tanθ²                 )      Tanθ        )  Sinθ/Cosθ  =

 

( __ 1/Sinθ dθ__ = (    CSCθ dθ 

  )                          )                        =Eew Gross!-- use a table,  or magically pull out of your butt the trick of timesing by  __ Cscθ +Cotθ___  

                                        Cscθ + Cotθ 

and using a U substitution where U = the denominator an dU = the numerator. Where the Heck did that come from?  Don’t worry about it, use a table.  Anyway when the dust settles integral solves to

= - ln |csc x + cot x| + C 

Solution Courtesy of http://www.math.com/tables/integrals/more/csc.htm

Note:  It might also be easier to solve Integral with partial fractions.

I can also use the general strategy of letting x = Secθ to solve the more complicated looking integral: 

 (___ dx____

  ) x √(x² -1)

Again let x = Secθ, Sec^-1(x) = θ and dx = SecθTanθdθ.  Making the appropriate substitutions leaves.

(___SecθTanθdθ ____ Using the identity = Secθ² -1 = Tanθ²

  )  Secθ √( Secθ ² -1)

(___SecθTanθdθ _ = (_SecθTanθdθ _ = ∫dθ = θ + C = Sec^-1 (x) +C

 )  Secθ √( Tanθ²)      ) SecθTanθdθ

Finally, let’s explore 

(  ___dx___    This one looks most similar in form to the Sin²θ identity so let x = Sinθ

  )     1-x²

If X = Sinθ then θ = Sin^-1(x) and dx = Cosθ dθ replacing.

(__ Cosθ dθ_  The bottom has an identity I can use to simplify.  1 – Sin²θ = Cos² θ.  Let’s use it.

  )    1-Sinθ²

(__ Cosθ dθ_  =  (__ Cosθ dθ_   = ( __dθ__  = ∫ Secθ dθ = crud!  I have to do more magic tricks.

   )    1-Sinθ²          )     Cosθ²           )   Cosθ     

In this case, the magic trick I need to do is times by ___Secθ +Tanθ___ 

                                                                                Secθ + Tanθ

Where did that come from?  The same bag of fairy dust that let me solve the Cscθ integral we ran into earlier.  Any way, this integral is evaluated by letting U equaling the new bottom U = Secθ + Tanθ and du = Secθ Tanθ + Sec² θ my old integral turns into the new and improved version

∫ du/u = Ln| Secθ + Tan θ | + C Where θ = Sin^-1 (x) which leaves us the nasty looking

Ln| Sec(Sin^-1 (x) ) + Tan (Sin^-1 (x) ) | + C. 

We can draw a trangle and get it a bit more solved and nicer looking, but hey our teacher has a Ph.D in math we’ll let him do the algebra if he wants it.  For our purposes we are done.  But that integral looked horrible so let’s examine a variation of the previous one.

If instead our integral looked like…

  (___ dx____ We could make the same substitution as before let x = Sinθ

   )  √(1-x² )

Thus θ = Sin^_1(x) and dx = Cosθ dθ. The identity 1 – Sinθ² = Cosθ² applies And Voila…

( __ Cosθ dθ__ = ∫dθ =θ + C = Sin^-1 (x) +C

  )    Cosθ

Ah good now that we know

a.     

 (  (___dx___    = ArcTanx + C

)     x² + 1

   (___ dx____ = ArcSec x +C

   ) x √(x² -1)

And (___ dx____ = ArcSin x +C

        )  √(1-x² )

By similar logic we arrived at the variations above.  As a final NOTE: It is useful to remember the identity Cosθ² =(1/2)[ 1+ Cos(2θ)].  This is a great way of drooping a squared exponent to a more manageable linear cos term.  

Stoichiometry: Aka Units of Doom!

Greetings all,  I have decided to post my explanations to common questions I am asked about math, chemistry, physics and biology.  Today's post is about stoichiometry in other words, unit conversions.  This is single-handedly THE most important topic in all of chemistry.  If you can master unit conversions you have 3/4ths of chemistry and 1/3 of physics down.

Here are some questions to get your brain juices thinking.  The answers to these questions should clear up your questions.  It may seem a little out in left field but stay with me.

What is 3/3?  How about 12/12?  127/127?  435296/435296?

The answer to all of them is 1.

Now what is 1*3?  How about 1*5?  1* 437?

Does timesing by 1 change the number you are multiplying by?

No.  One is what is called the multiplicative identity.  Meaning I can times anything by one and it doesn't change the number, only what it looks like. 1*5 looks a little different than just 5 but it still = 5.  This is useful for chemistry because I can change the units of the quantity I am working with without changing the amount. By doing unit conversions I am really multiplying by a complicated version of the number one (The loneliest number).

Next question, am I the same person if I stand on my head?  Am I the same if I lie on my side?  In short does it matter what orientation I am in or am I still me even on my head?

Of course I am the same.  The same is true for conversion factors.  They still work even if they are upside down.

A few more things then we will get to the actual chemistry.


___________|_____________|____

________|_________
                   |                       |                     |

This is what I like to call "railroad tracks."  Its a table that I will use to keep track of what units I am in an what units I am converting into.  This is one of the most important tables in all of chemistry (the other is the periodic table.)  This little table lets me know instantly if I got a problem right or not.  This is one reason I love chemistry it is one of the few things with instant feedback for a correct response to a tricky puzzle.  If you can master this table you have most of chemistry mastered.  For at least the first 2.5 years of chemistry all it really is is a series of fancy unit conversions.

Lets practice one problem using this table to make sure we get the hang of it (Note this table can be as long or as short as I need it depending on the problem.). Suppose you are given the problem, "How many seconds are in 45 years?"  How on Earth do you begin?  Simple. Search the problem for a number that has only 1 unit attached to it.  In this case ___ years.  Years is only one unit  If I had something like Kg m/s^2 that is four units all lumped together, one usually doesn't start with that.  There are some exceptions where it is actually easier to start with the messiest looking thing, but usually start with the given that has one unit.  In this case years.  The thing you are starting with goes in the top left square of our rail road tracks.

__45 years_|_____________|____________|_________
                  |                       |                     |

What is a conversion factor you know involving years?
Oh I know ~365 days/year.

Ok so where do I put it?  Well I want my years to cancel so it must go on the bottom.


__45 years_|_365 days___|____________|_________
                  |    1 years     |                     |

Note: "per" means put it on the bottom.  Miles PER hour = __miles__
                                                                                          hour

If I did this WRONG my units would not cancel.  It would look like...  

__45 years_|_1 years______|____________|_________
                  |   365 days      |                     |

And I would be left with units of Years ^2/day.  And I ask you, what the heck is a square year per day!?  What does that even mean.  Seeing the above I would immediately recognize Ooops !  I did something wrong.  Seeing this its a simple fix to erase the part that is wrong and flip it over.  In fact if i can't think of anything else to do start converting units stop when the units are what I want.  If you don't get full credit for this you will at least get quite a bit of partial credit.  So lets continue. What is a conversion factor involving days? 24 hours/day.  Plug in appropriately and you get...

__45 years_|_365 days___|___24 hours_____|_________
                  |    1 years     |       1 day            |
                                                                      
Continue with 60s= 1 min and 60 min = 1 hr. And your final table looks something like this...

_45 years_|_365 days___|___24 hours_____|___60 min__|_60 sec___
                  |    1 years    |       1 day           |      1 hr       |  1 min

Aha!  I am not in units of secconds, the units I wanted.  To make sure I did every thing right, I double check to make sure all of the others canceled.  Years divided by years, days divided by days, hours/hr & min/min.  Sure enough they all canceled.  I am left with seconds, which is the units I wanted all along.  Ok great!  Now what?  What do I do with this table.  Times everything that is on top by each other (just the numbers I already got rid of the units.)  And divide by everything on the bottom.  When I enter this into my calculator I should enter it as follows:

"(45*365*24*60*60) / (1*1*1*1) = "

Note:  The parenthesis are VERY VERY important. In our case the bottom was just 1*1 and we can do that in our head.  However, you are not always so lucky to have ones on the bottom.  Sometimes you will have numbers like 6.022*10^23 or other giant beasts lurking in your basement.  If you do not put parenthesis your calculator will take one number divide it by another and then multiply it by your bottom.  I actually saw someone make this mistake.

__3 Units A__|__ 5 Unit B___|__ 1 Unit C__
                     |     2 Units A   |  6.022*10^23 Units B
           
He entered it into his calculator as:

3*5*1 / 2*6.022*10^23

The calculator interpreted this as...
3*5*1*6.022*10^23 / 2

So because he didn't use parenthesis my tutee was off by 45 orders of magnitude!
How he should have entered it is like so...

"(3*5*1) / (2*6.022*10^23)"

That way the calculator will divide by 1.2044*10^24 instead of dividing by 2.  Big difference!  USE parenthesis!

Now to the questions at hand (From this point forward you will need a periodic table--go get one!)

1) Determine the mass of Sodium hydroxide produced when (whenever you see a decimal point put a 0 in front of it, it is harder to loose track of 0.~~ than .~~.  You are less likely to think it is a random pencil squiggle than a decimal point if you include the 0.  So always include a 0) 0.47 g of Sodium nitride reacts with water according to the following equation:    Na(3)N + 3H(2)O ---- NH(3) + 3NaOH

Step 1. Set up the railroad tracks.

___________|_____________|____________|_________
                   |                       |                     |

Step 2. Find what is given with only 1 unit.

0.47g NA(3)N Include the name of what you are playing with.  It is quite common in chemistry to go from moles of A to moles of B so include the As and Bs to avoid getting lost.

Step 3. Fill in the table.

___0.47 g Na3N___|_____________|____________|_________
                            |                       |                     |

How much does 1 mole of Na3N weigh?  Pull out your periodic table look at the numbers on the bottom add them together i.e. 3*23.99 for the 3 sodiums +14.01 for the Nitrogen for a total weight of 85.98 amu, but what the heck is an AMU?  Is it some sort of guns store?  No it stands for grams/mole.  Hey we have grams on our chart  lets plug in the weight and convert it to moles.  But where does it go, on top or on the bottom?  Where must we put it so that grams will cancel?  On the bottom of course.

___0.47 g Na3N___|____1 mole_Na3N__ |____________|_________
                            |      85.98  g              |                     |

Now where do I find the ratio of sodium nitride to sodium hydroxide?  If only there was something that told me how much Na3N and NaOH would react together.  If only I had some sort of equation relating all my reactants and products telling me how much of each I would need.  If only I had such an equation I would be set.  Oh I do have such an equation.  For every 1 mole of Na3N I produce 3 moles of NaOH.  Incidently the numbers of the balanced reaction coefficients work for volumes and numbers of atoms but not for mass.  Consider this:  I weigh 200 pounds.  In this room right now there are 600 pounds.  Does this mean there are three of me?  No.  So balanced reactions do not work for mass, only volumes and number of atoms.

___0.47 g Na3N___|____1 mole_Na3N__ |___3 mole NaOH__|_40 g________
                            |      85.98  g              |     1 mole Na3N    | 1 mole NaOH

Final step find the mass of sodium hydroxide. 23+16+1.

Times it out and you get...

(0.47*1*3*40)/(85.98*1*1) = your answer.  Don't forget the parenthesis!

For Mass-mass use the periodic table.  For mass-volume conversions you will need to use the density (mass/volume) or the concentration (moles/liter).  For Volume-Volume use the coefficients of the balanced reaction (or possibly the ideal gas law if it is a gas.).

For limiting reactants you have to actually do the problem 2x.  Once assuming one of the reagents is the limiting and again assuming the other is the limitant. The trick is to start with a balanced reaction and to pick one substance and convert everything into that one substance. So for example 2. I would set it up as follows...

1. Balance the reaction

__O2 +__H2 = __H2O

When I do this I get...

1-O2 + 2-H2 = 2-H2O

2. Plug in numbers

___2.41 g O2__(From Question)_|_1 mole O2 (From Reaction)_______ |_2 mole H2O (From Rxn)
                                                 |  32.00  g O2 (From Periodic Table)   | 2 mole H2

= A moles of Water

___3.67 g H2___|____1 mole H2___ |___2 moles H2O__ = B moles Water
                         |      2.02  g H2       |     2 mole H2       

Compare A and B see which one is smaller.  The one that ones out first is the limiting reactant.  So if A is smaller than B.  Oxygen must run out first and therefore oxygen would be the limiting reactant.

% yield is just how much I got divided by how much I should have got *100 to make it a percent.

Usually they will give you how much you actually got and how much you started with.  Use the amount you started with to calculuate how much you should have had if you did it perfectly.

Consider the reaction

C + 2-H2 explodes and makes CH4.  I put explodes because it sounded cooler. It doesn't really.
Suppose you started with 3.0 g H and you produced 2.7 grams C what is your % yield.

If I did this perfectly all of H should react to produce CH4.  Using a periodic table I find H2 weighs 2.02 g/mole. While CH4 weighs 16.05 g/mole.

___3.0 g H2___|_1 mole H2___  |___1 mole CH4  _|_ 16.05 g____| = 11.9 g CH4 
                       |   2.02 g H2       |     2moles H2     | 1 mole CH4  |

Wait how can my theoretical mass be larger than the mass of hydrogen I started with?  I added carbon as well. Oh yea. Though my final mass should never be larger that the sum of both parts I added.  Carbon AND Hydrogen.  If it is I did something wrong.

So If I did it perfect i should haved produced 11.9 g CH4 I actually produced 2.7g.
%Yield = actual yield / Theoretical Yield or real world/ perfect world
 => (2.7 / 11.9)*100 = 22.9%  Yea! My reaction lost 87.1% of the stuff I put into it.  Meaning I spilled most of my beaker.  If my percent yield is ever greater than 100% I probably did something wrong.