Excess & Limiting Reactants persent yield

A.Excess Quantity
A balanced equation describes what should happen in a chemical reaction.

However, the conditions necessary for the reaction to take place may not be present. (ie. pressure, temperature, concentration etc.)

Sometimes it is necessary to add more of one reactant than the equation predicts because it is not possible for every atom/molecule of the reactants to come together.
Ex. One reactant is the Excess Quantity and some of it will be left over, the second reactant is used up completely, and is the limiting reactant.

B.Excess quantities in chemical reactions.
Ex1. How many grams of OCl2 will be formed when 44.0 g of O2 reactant with 97.0g of Cl2

(*General Process: Convert both reactants to the desired product & the smaller amount of product will actually be produced.)

Step1: Balanced equation.
O2(g) +2 Cl2(g) → 2OCl2(g) 

Step2: Convert Cl2 to OCl2
97.0 g ×1mol Cl2/71g × 2mol OCl2/2mol Cl2 × 87.0g/1mol OCl = 118.86 g OCl2

Step3: Convert O2 to OCl2
44.0g × 1molO2/32g ×2mol OCl2/1mol O2 × 87.0g/1mol OCl = 239.25g OCl2
Because 97.0g of Cl2 reacts to produce the smaller amount of product. Cl2 is limiting quantity and O2 is excess quantity. 119g is the mass of product.

Ex2. 
41g O2 reacts of 164g of Cl2, Which is EXCESS quantity? Which is Limiting quantity? How many grams of the excess quantity will be used?
Step1:
O2(g) +2 Cl2(g) → 2OCl2(g) 
41.0g       164g
(*General process: Convert ONE of products to the other reactant to see which excess and which is limiting. Determine which is left over and by how much. It may involve another conversation)
Step2:
To find which reatant is in excess calculate how many grams of Cl2 gas would be required to react with 41.0g of O2.
41g O2 × 1mol O2/32.0gO2 × 2mol Cl2/1mol H2 × 71.0g Cl2/1mol Cl2= 181.93g=182g Cl2

Since there are only 164g of Cl2 gas, not all O2 can react. Thus Cl2 is limiting quantity and O2 is excess quatity.

Step3: Convert Cl2 to O2 to see how much O2 would be needed to react with 164g of Cl2.
164g Cl2 × 1mol Cl2/71.0g Cl2 × 1mol O2/2mol Cl2 × 32.0g/1mol O2 = 36.958=37.0 g O2

37.0g of O2 would react with 164g of Cl2.

STOICHIOMETRY (2) !!

Example 1 :
a) How many grams of Fe(Ⅱ) will be needed to react with 3.20 mol of HCl?

Step1: Balance the equation.
1Fe + 2HCl → 1FeCl2 + 1H2

Step2: Make a "ROAD MAP"

Step 3: Do calculation

3.2 mol HCl × 1mol Fe/ 2mol HCl × 56g Fe/ 1mol Fe = 90g Fe

b) How many grams of Fe will be needed to produce 10.0 g H2

10.0 g × 1mol/2.0 g H2 × 1mol Fe/ 1mol H2 × 56 g Fe/1 mol Fe =280 g Fe

Note: grams (X) ↔ mol(x) ↔ mol(y) ↔ grams(y)

Example 2 :

a) Using the following equation : 2 NaOH + 1H2SO4 → 2H2O + 1Na2SO4

How many grams of sodium sulphate will be formed if you start with 200 grams of sodium hydroxide?

200 g NaOH × 1mol NaOH/ 40 g × 1mol Na2SO4/2 mol NaOH × 355.3g Na2SO4/ 1mol = 888.25 g

                                                                                                                                            = 900 g

STOICHIOMETRY (1) !!

What is stoichiometry?
stochio=Greek for element
metry=measurement

Stoichiometry is a branch of chemistry that deals with the quantitative analysis of chemical reactions and is about measuring the amounts of the elements and compounds involved in a reaction.

It is the study of the relationship between the amount of reactants uesd in a chemical reaction and the amounts of products produced by the reaction.

Consider the chemical reaction:
4NH3+5O2→6H2O=4NO
with stoichiometry, we find out:
4:5:6:4
do more than just multiply atoms
4:5:6:4
are what we call a mole ratio.

Balanced chemical equations are required.
It tells us the ratio of the molecules or moles of the substances in a chemical reaction.

ex
For the equation: Zn(s)+2HCl(aq)→ZnCl2(aq)+H2(g)
a) How many atomes of Zn are needed to produce 1 molecule of Hydrogen?
1 molecule H×1atom Zn/1molecule H=1atom Zn

b)How many moles of HCl are needed to produce 0.452 moles of Zinc chloride?
0.452mole ZnCl2×2mole HCl/1mole ZnCl2=0.904mole HCl

c)How many grams of Zn will react with 1.05 moles of HCl?
1.05mole HCl×1mole Zn/2mole HCl=0.525mole Zn  0.525mole Zn×65.4g/1mole=0.038mole H

 

The Calculation of Energy

Energy Calculation

--the return of mole

CH₄+2O₂àCO₂+2H₂O

△H for this exothermic reaction is expressed using the coefficients of the balanced equation:

-812KJ/1mol CH₄ or -812KJ/2mol O₂ = -406KJ/1mol O₂

So the value of △H depends on which chemical you are referring to

1.    The value of △H changes with different reactions; -812KJ is only specific to this reaction. Thus like molar mass is different for each compound, △H depends on the chemical reaction

2.    △H is not a constant like Avogadro’s number

Ex. How many moles of CH₄ are needed to produce 2100KJ of energy?

CH₄+2O₂àCO₂+2H₂O+812KJ

-2100KJ×-1mol CH₄/812KJ =2.6mol

TYPES OF REACTIONS (2)

Double Replacement
-A double replacement is a reaction between two IONIC compounds usually in solution. The ions switch partners like a dance. The (+) ions sitch places but remember to watch your charges.

General Formula
AB + CD→ CB+AD

Eg: CaCO3 + 2HCl → CaCl2 +  2H2

If a double replacement reaction actually occurs:

The reactants change state during reaction

(usually a precipitate occurring)

Use your "Table of solubilities" to determine the states.- (aq) or (s)

Net Ionic Equation:

-There is a net reaction when you have a precipitation that occurs

-(aq) ions that are the same on both sides get cancelled

Eg.

CaCl2 + Na2CO3 → CaCO3 + 2 NaCl

Net Ionic Equation： 2Cl^-  + 2Na⁺  → 2NaCl

Using the Table of Solubilities

1) Find your anion ( negative ion) in the left hand column.

2) Look for your cation (positive ion) in the list in the 2nd column.

3) Follow its presence or absence to the word "soluble" or "not soluble"

4) If soluble, the compound is (aq)

5) If insoluble, the compound is (s)

Combustion

-A combusion reaction is a reaction where burning in air is involved. The reactants are the chemical to be burned and the oxygen that it reats with. The oxygen atoms usually end up combined with more that one type of atom as products.

-General Formula

AB + O2 → AO + BO

Eg. CH4 + 2 O2 → 2 H2O + CO2

Neutralization

-A neutralization reaction is a sepcial double replacement reaction where acids reat with bases to produce water and an ionic salt as products.

-The acids have han H+ as the cation (+) and the bases have  OH- as the anion (-). Both should be aqueous solutions. (aq)

-General Formula

HA + BOH → H2O + BA

Eg. HCl + NaOH → NaCl + H2O

NOTE： For DR reaction (include Neutralization）, some ions participate in the reaction wile other ions do not participate.

Eg. KCl(aq） + AgNO3(aq) → KNO3(aq) + AgCl(s)

* Be sure to write both the total and net ionic equations!!!

Total  ionic equation:  K⁺_(aq)   + Cl^- _(aq) + Ag⁺_(aq) + NO₃^-_(aq） → K⁺_(aq)  + NO₃^-_(aq) + AgCl_(s)

Net Equation: Ag⁺_(aq) + Cl^-_(aq)  → AgCl_(s)

Endothermic And Exothermic Reactions

Introduction
-All chemical reactions involves △Energy
Some release energy.  exothermic
Some absorb energy. endothermic


Molecules are held together by chemical bonds.
-Add energy to break bonds
-Give off energy to join together

Takes MORE enerygy to break down than it gives off to form bonds→Endothermic

Takes LESS energy to break down than it gives off to form bonds→ Exothermic

Enthalpy, H, heat contained in the system.

Energy Diagrams
-Chart the potential energy of the chemicals as they change from reactants to products
-Reactants start with a certain amount of energym energy is added to start the reactions and then energy is released as the reaction proceeds.
-The relative amounts of energy determine: Endothermic & Exothermic.

Energy of reactants: total E of all reactants in the reaction
Energy of products: total E of all products in the reaction
Energy of activated complex: potentioal E of the "transition state" between reactants & products
Activation Energy: The E must be added to get the reaction to progress.
△H: the change in potential energy during the reaction.
△H= E of reactants - E of products

Potential Energy Diagram

E of reactants ＜ E of products                                               E of reactants  ＞E of products

△H is positive.                                                                        △H is negative.

Endothermic reaction                                                               Exothermic reaction.

The Energy absorption or release can be placed directly in the equation.

Eg. CH4+2O2→CO2+2H2O+812KJ

      Higher E        Lower E

Exothermic reactions have the E term on the RIGHT side(RHS) -△H

Endothermic reactions have the E term on the Left side (LHS)    +△H

Eg.

E of Reactants= 150 KJ

E of Products= 100KJ

E of Activated complex= 300 KJ

△H=?      Ea=?

△H= E products- E reactants

      = 100KJ - 150KJ= -50KJ  ( exothermic)

Ea= E activated complex - E reactants

    = 300KJ - 150KJ= 150KJ

Eg.

E of Reactants （E）=200 KJ
△H（F）                =100 KJ ( Endothermic)
Ea(B)                        =300 KJ
E of activated complex (C) =?      E of Product (G)=?      

E of activated complex= Ea+ E of reactants= 200KJ+300KJ=500KJ
E of product= E of activated complex- E of reactants= 500 KJ- 200KJ= 300KJ

Lab 5B

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

Material and Equipment:

Refer to page 33 in Health Lab Text, Lab 5B

Procedure:

  Refer to page 33 in Health Lab Text, Lab 5B

Data and Observation:

  Attached to the back.

Analysis of results:

1.       Copper reacts with oxygen in the air.

2.       The solution become lighter blue indicates the concentration of CuSO4 in the solution decreased.

3.       The colour changes in reaction 3&4, which means new substance was yielded.

4.       The precipitate was CaCO3.

5.       Put a glowing (not burning) splint into the mouth of the test tube, the splint will start a flame. That shows the identity of supporting combustion.

6.       A. has the identity of supporting combustion.

B. Water.

Follow-up questions

1.       CdSO4+ZnàZnSO4+Cd

2.       2H2Oà2H2+O2 (decomposition)

Conclusion

1.       2Cu+O2à2CuO (synthesis)

2.       2Fe+3CuSO4àFe2(SO4)3+3Cu (single replacement)

3.       CuSO4·5H2OàCuSO4+5H2O  (decomposition)

4.       CuSO4+5H2Oà CuSO4·5H2O (synthesis)

5.       CaCl2+Na2CO3àCaCO3+2NaCl (double replacement)

6.       Zn+2HClàZnCl2+H2 (single replacement)

7.       2H2O2+(MnO2)à2H2+O2+(MnO2*) (decomposition)  (*=catalyst)