Lab 6D- Determining the Limiting Reactant and Percent Yield in a Precipitation Reaction

Alrighty! So, why are we doing this lab?

We, as diligent chemistry students, want to find out the reaction between the solutions NaCO3 and CaCl2, determine which is limiting and which is excess reactant, determine what the theoretical mass of the precipitate that should form, AND to compare the actual mass to the theoretical mass of the precipitate and calculate the percent yield.

In order to do this lab you will need...

• SAFETY GOGGLES. (Here's the thing, it might shock you... deep breath... YOU MIGHT NEED YOUR EYES IN THE FUTURE. LE GASP!)
• 1 Centigram
• 2 gradulated cylinders (25mL)
• 1 beaker (250mL)
• 1 wash bottle
• 1 filtering apparatus (1 ring stand, Erlenmeyer flask (250mL), 1 funnel)
• 1 filter paper
• 25mL of 0.70M NaCO3 solution
• 25mL of 0.50M CaCl2 solution

Now, after you've got all your things, all you have to do is...

1. COMBINE the NaCO3 solution with the CaCl2 solution
2. Observe... wait now for 5 minutes. You can weigh your filter paper.
3. While waiting for the reaction to fully react, set up your filtering apparatus.
4. When setting up your filtering apparatus, using the wash bottle, lightly wet the filter paper so it sticks to the funnel.
5. Swirl your beaker of precipitation and slowly pour your solution into the apparatus. Take your time with this.
6. Using the wash bottle, squirt some water into the beaker to get the remaining precipitate out. Also rinse the precipitate in the filter paper to remove the NaCl2.
7. Remove the filter paper and let it dry.
8. Clean up.

Day 2

1. After it has dried, weigh it again. Record.

Now, wasn't that easy? And also SOOOO fun? Do you know what this white precipitate is? It is actually.... *drum roll*

CHALK?! (Don't worry it won't kill you... BUT THAT DOESN'T MEAN YOU SHOULD EAT IT.)

The equation for this bad boy...

1 Na2CO3 + 1 CaCl2 --> 2 NaCl + 1 CaCo3

Have Fun!!

Percent Yield and Percent Purity

Percent Yield
In chemical reaction, products are always not the same as in calculation. For example, in stoichiometry calculation, we are supposed to get 12g CaCO3 in this reaction: Ca(OH)2 + H2SO3 ------CaSO3 + 2H2O. But in fact we just get 8g. The reason might be impure reactants. Therefore, we have Percent Yield.


Percent Yield is the percentage of real mass we get by the calculation results.

% Yield = actual mass / calculation mass

Percent Purity
Same as the other substance. Sometimes the substance might not be absolute pure. For example, gold alwas contains impurities in it.

% Purity = pure mass / whole mass

Excess and Limiting quantity

so we've learned a new way that most of the reactions accur don't use up all of the reactant but will be some left over.
For example when i have 40 speakers and 10 mics. Provided that to make a head set, you need 2 speakers and 1 mic. So for the amount I have before, i can only make 10 headsets right? And of course there will be 20 speakers left.
So does in the reaction, usually it's not used all of the amount of reactant to form product. That's why today we learn how to calculating that left over.

During my review for the HORRIBLE TEST, i just realized a very helpful strategy to solve this problem without any concern. HERE YOU GOOOO:

So i'm gonna get a problem in the review sheet.
If 15 grams of Cu (II) Chloride react with 20 grams of NaNO3
a/ Write balanced equation (c'mon EASIIIIIIIIIIIII)
b/ What is the limiting reagent
c/ How much NaCl can be formed ?

I BELIEVE THAT WHEN YOU SOLVE THIS PROBLEM PRECISELY, YOU’RE GONNA BE ABLE TO SOLVE ALL THE OTHER PROBLEMS.

CuCl2 + 2 NaNO3 à Cu(NO3)2 + 2 NaCl

Mole in equation: 1 2 1 2

Mole actually reacted 0,1 0,2 0,1 0,2

So basically you calculate the mole from the question. Because they’ve given you the mass of both reactants so you can calculate it.

Mole CuCl2 : 15/135 = 0,1 moles

Mole NaNO3: 20/85 = 0,24 moles

After this, you would have to compare between 2 moles above by getting the ratio between that and the mole in equation.

So: 0,1/1 < 0,24/2 => the theorical mole of NaNO3 is larger than CuCl2 à we can easily say that CuCl2 is the limiting quantity.

From this, please forget about the mole of excess (NaNO3) cause we can only use the limiting moles to calculate thing. Plug it into your equation and calculate the other moles by their coefficient.

c/ in order to calculate the excess mole, you just get the mole just calculated from the given information (0,24) subtract by the mole that actually reacted (0,2) then you can get 0,04 mole of left over.

The secrets of STOICHIOMETRY

Hello there.  You may be wondering, what Stoichiometry means.  Well:

Stoichiometry is the quantitative analysis of chemical reactions and is about measuring the amounts of substances involved in a reaction.


What??  You don't know what this means??  Well, to put it into more simple english, it is the study between the amount of reactants used in a reaction and the amounts of a products produced by a reaction.

In more simpler english, it is the study of ratios between reactants and products.


Here is a sample question of stoichiometry:

For the equation: Zn + 2HCl ---> ZnCl2 + H2

1) How many atoms of Zn are needed to produce 1 molecule of Hydrogen?
Answer: 1, because the ratio of Zn to Hydrogen is 1:1, therefore the answer is 1.


2) How many grams of Zinc Chloride will be formed when 2.00g of Hydrogen is formed?

Step 1 - Convert 2.00g of Hydrogen to moles

2.00g x 1 mol/2.0g = 1 mole of Hydrogen.

Step 2 - Plot the mole into the mole ratio.

In this case, the ratio is 1 mol ZnCl2 : 1 mol H2, because there is one mole of hydrogen.

Step 3 - Answer the question, in this case, convert the moles back into mass.

1 mol ZnCl2 x 136.4g/1mol = 136.4 g of ZnCl2

136.4g of ZnCl2 will be produced when 2.00 g of H is formed.


Still don't understand?  Watch the video to learn more!

Balancing Equation

Here is an equation

K+O2 ---K2O
It's a skeleton equation. It showed what the reactants and products are. But not ALL the atoms. On the left, it has one K, two O. But on the right, It has two K and only one O. Therefore we need to balance it.

A balanced chemical equation shows all atoms.

Because on the left it has two O. so we need to add up the number of O on the right hand side. Write down 2 before K2O. And we have 4 K and 2 O. balance the K. Write down 4 before K on the left hand side.

Here we have a balanced equation.     4K+O2 ---- 2K2O

Some tips for balancing equation

1. draw a chart and write all the numbers of atoms, that makes faster and easier calculation
2. assume the most complicate compounds as 1
3. always double check after balancing

Translating Word Equation/naming compounds

Translating Word Equation.

There are two ways to write an equation, word equation and symbolic equation.
Word equation contains reactants and products, no need for balancing.

Example:
This is a symbolic equation 2NO+O2-----2NO2
change it into word equation:
Nitrogen monooxide+oxygen-------nitrogen dioxide.

1. name the compound names first.
2. write down the equation.

Naming compounds

two types of compounds: ionic and covalent
   
 Ionic compound

1.  no prefix needed. But have to drop the end of negative ion and add -ide
2. Roman numerals are used following the positive ion to indicate which ion was used.

Example: manganese(III) sulfide                 ---- Mn2S3


Covalent compounds

1. Prefix is needed. (1-mono 2-di 3-tri 4-tetra 5-penta 6-hexa 7-hepta 8-octa 9-nona 10-deca)
2. drop the end of negative ion and add -ide

Example: N2O4         ----------dinitrogen tetraoxide

Pre-Lab 5B Types of Chemical Reactions

Pre-Lab Types of Chemical Reactions


Objectives
1)  To observe a variety of chemical reactions.
2)  To interpret and explain observations with balanced chemical equations.
3)  To classify each reaction as one of the four main types.

Supplies

Equipment

lab burner
6 test tubes
      one will be flame heated
test-tube clamp
medicine dropper
wood splints
cruicible tongs
steel wool
safety goggles
lab apron

Chemical Reagents
copper wire(bare)
iron nail
0.5M copper(II) sulfate solution
solid copper (II) sulfate pentahydrate
water
0.5M calcium chloride solution
0.5M sodium carbonate solution
mossy zinc
2M hydrochloric acid solution
hydrogen peroxide solution(6%)
manganese (IV) oxide


Procedure

1)Put on lab coat and goggles
2)Observe before, during , and after each reaction.  Record in table one.
Reaction one
3) Adjust burner flame to high heat

4)Using crucible tongs, hold 6 cm length bare copper over hottest part of flame for a few minutes.
Reaction Two
5) Clean iron nail with steel wool until it turns shiny
6) Nail in tube and copper(II) sulfate solution so half of nail is covered.
7) After 15 minutes, remove nail and note changes








Reaction Three
8) Put solid copper(II) sulfate penta hydrate until the tube is one third full.
9) With clamp, angle away from peers and heat (back and forth gently)
10) Continue until no more changes.

             





Reaction Four
11) Allow contents from reaction three to cool.
12) Use medicine dropper to add two or three drops water to tube.

Reaction Five
13) Fill a tube a quarter full with calcium chloride solution.  Fill second tube 1/4 full sodium carbonate solution.
14) Pour calcium chloride into the sodium carbonate solution.





Reaction Six
15)Place a piece of mossy zinc into the test tube.
16) Add HCl into the tube until the zinc is completely covered

Reaction Seven
17) Half fill a tube with hydroperoxide.
18) Add small amount of manganese (III) oxide.
19) Test gas by placing glow(NOT BURNING) splint into the mouth of a tube.

20) Clean-up!