Lab 6D

Today we did a super cool expriment on Limiting Reactant and Percent Yield in a precipitation reaction.  Our objective is ofcourse to first observe the reaction in the double replace ment reaction
Na2Co39(aq)+CaCl2(aq)------->NaCl(aq) + CaCO3(S).  Then we determine the limiting reactant and the excess reactant.  At the end will will than classify the Percent Yield by comparing our product produce in this experiment and the theoritical product.

Procedure:
1.  In the begining of the lab we put on our safety equipments.
2. We obtain Na2Co3 solutions and CaCl2 solution and record our measurements.
3. Than we poor both sultions into a beaker and leave it sit for 5min
4. Than we write our name on a filter paper and set up the filter apparatus like this

5. We then place grducated cylinder underneath the funnel to start filtering. 
6.  We will use a wash bottle to rinse any NaCl left on the filter paper so that only the precipatate remains.
7. At the end we remove the filter paper with the precipitate CaCO3 into a dry location so that we can measure the percent yiled next class.

End of the Lab
With the double replacement balanced equation  1Na2CO3(aq)+1CaCl2(aq)------->1NaCl(aq) + 1CaCO3(S) we can find the limiting reactant. To find the limiting reactant, we simply just convert one reactant to another to see if they exceed more they are presented or the Presented is more than it is needed

E.g  we use (25ml) of 0.70M Na2CO3 and (25ml) of 0.50M of CaCl2.  In the end we converted them into moles.  0.018mol Na2CO3 and 0.012mol CaCl2.

0.0125molCaCl2* molNaCO3/molCaCl2=0.0125mol CaCl2.  Thus CaCl2 is the limiting since it needs more than it is given.

Conclusion
To find the percent yield of this experiment, we will need to weigh the mass of the filter product which is next class.  The formula to find the percent yied is

Percent Yield=   actual mass produced (grams)    x 100

           theoretical mass produced (grams)

Stoichiometry: Excess and Limiting Reactants Percent Yield

In this section, we learned that for a balanced equation, there is some conditions that the reaction may not present ( may be caused by pressure, temperature, concentration, etc.). Sometimes it is necessary to add more reactant into the chemical reaction in order to react.

One reactant is the EXCESS QUANTITY that would have some left overs at the end. The second reactant is used up is called the LIMITING QUANTITY.

The videos belows tells you how to calculate excess quantities and limiting quantities based on the equation.
ENJOY!

Excess and Limiting Reactants

Today in class we learned the reactants in a chemical reaction is not always exact which means some reactant may have more amount than others.  Thus there will be excess of reactants during the reaction.  We also learned about the limiting reactant; a reactant that is not present in alarge enough quantity to fully react with another reactant.  The limiting reactant is significant because it helps us determine how much product can be formed.  This concludes what we learned today.

Examples





In this cass the car bodies are the litmiting reactant becaucause not matter how many tires there are only 8 car bodies are availible to make a car.  Thus only 8 cars can be formed and there is 16 tires excess.



A sample of 111.6g of Fe is mixed with 96.3g of  S.

a) Which reactant is in excess

b) Which is the limiting reactant?
c) When this reaction is carried out, what mass of FeS will actually be produced?

Solution
1) Write out the balance equation.
1Fe+1S----->1FeS

2)Convert Fe and S into moles
Mol Fe= 111.6g Fe *1molFe/55.8gFe =2.00 mol Fe
Mol S= 96.3g S * 1molS/32.1gS= 3.00mol S

3) Use mol ration to determined how much S is need to react with Fe
2.00mol Fe* 1molS/1molFe= 2.00mol S needed.  Thus S is the excess reactant because 3.00 moles of S is presented in the beginning. Since S is the excess than Fe will be the litming reactant.

4)Use the mass of the limiting reactant to find the mass of FeS that actually produced because we cannot start with the mass of the excess reactant S.  We need to use Fe the limiting reactant or we will get more mass of the product than we actually can.
111.6g Fe* 1molFe/55.8g Fe*1molFeS/1molFe*87.9gFeS/1molFeS=176gFeS

Molarity and Stoichiometry

So, today we've taken stoichiometry one step further, which is including the molarity.

Remember the stuff we did last day on stoichiometry was basically to take gram stuff into the mole stuff and such and such, but today, we involved molarity.

As we know, molarity is a concentration stuff, so for example:

-How many mL of 0.124M NaOH contain enough NaOH to react with 15.4 mL of 0.108 M H2SO4 ?

-First of all, we have to wrtie the chemical reaction out and balance it.

    #mL       15.4mL
2 NaOH + H2SO4 ---> 2 H2O + Na2SO4
0.124 mol   0.108 mol

--------------- --------------

     1L               1L

and the question has given out a lot of information, as you can see, it is asking the milliliter of NaOH (in red). And we all know the molarity can be written as moles per L, so the question said 0.124 M, which is equivelent to 0.124 mole per L (in yellow). and another piece of information is 15.4 mL (in green) of 0.108M, or 0.108 mole per L (in aqua).

And now let's start doing some conversion!!


 15.4 mL           1 L          0.108 mol H2SO4     2 mol NaOH     
-------------- x --------------- x ----------------------------- x ---------------------
   1 L           1000 mL                   1 L              1 mol H2SO4   
                  1 L             1000mL
    x ------------------------- x ------------ = 26.82580645 mL NaOH   <---- WRONG!
       0.124 mol NaOH        1 L

Be careful !! sig fig counts for mark!!

26.82580645 mL NaOH, 3 sig figs = 26.8 mL NaOH

+

 

YAY ! 




Stochiometry Involving Moles

        

Today in Class we learnt how to calculate from moles to grams and vice versa.  The  balance equation shows the relationship between moles of the substances nicely.  However in real experiments we deal with the mass of the substances which is in grams.  Therefore we need to know how to convert one way to the other.

Review

How to find the mass of substance B when you are given the mass of substance A?

Solution:  A(grams)--> moles of A-->moles of b--->mass of B(grams)






Road map for all calculation involving Stoichiometry

 



Practice Web#1


 Practice Web#2



Road map focusing on grams to moles

 

Example:

WHAT IS THE MASS OF O2 IS REQUIRED TO REACT WITH 80.0 GRAMS OF CH4?

CH4 + 2O2--> CO2 + 2H2O



Step 1:  Check if the equation is balance.

 Solution:  Each side has 4H, 4O, and 1C.  Thus the equation is balanced.

Step 2:  Convert 80g of CH4 to moles of  CH4

Solution: 80gCH4*1molCH4/16.0gCH4= 5molCH4

Step3:  Convert 5molCH4 to molO2

Solution: 5molCH4*2molO2/ 1molCH4= 10mol of O2

Step 4:  Convert 10molO2 to O2g

Solution: 2.5mol02*32gO2/1molO2= 320g02

Step5:  Check your Sig Figs

Solution: Number use in the quetion was 80.0 so that is three sig figs.  Thus 320g0--->3.20*10^3g of O.



Stoichiometry

Stoichiometry is a branch of chemistry that deals with the quantitative relationships that exist between the reactants and products in chemical reactions. In a balanced chemical reaction, the relations among quantities of reactants and products typically form a ratio of whole numbers. For example, in a reaction that forms ammonia (NH₃), exactly one molecule of nitrogen (N₂) reacts with three molecules of hydrogen (H₂) to produce two molecules of NH₃:

N₂ + 3H₂ → 2NH₃

Stoichiometry can be used to calculate quantities such as the amount of products that can be produced with given reactants and percent yield (the percentage of the given reactant that is made into the product). Stoichiometry calculations can predict how elements and components diluted in a standard solution react in experimental conditions. Stoichiometry is founded on the law of conservation of mass: the mass of the reactants equals the mass of the products.

Reaction Stoichiometry describes the quantitative relationships among substances as they participate in chemical reactions. In the example above, reaction stoichiometry describes the 1:3:2 ratio of molecules of nitrogen, hydrogen, and ammonia.

Remember:
-balanced the chemical equation before solving the problem.
-what you need over what you have, where you're going over where you from.

Example:

     Consider the chemical equation:

           N₂ + 3H₂ → 2NH₃

 Q: How many moles of  H_{2 will be formed when 3.7 moles of} NH_{3 is decomposed?}

                                              

                                           _{3 moles of}  H₂

             _{3.7 moles of}  NH_{3  x  ____________  =  5.55 moles of} H₂ 

                                                    

                                                         _{2 moles of} NH₃