101 lab lecture 3 effect of limiting the concentration of

8/13/2019 101 Lab Lecture 3 Effect of Limiting the Concentration Of

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Purpose1. To determine the limiting reactant in a salt

mixture2. To observe the effectof a limiting reactant onthe amount of product formed using differentconcentrations of starting material

3. To compare the different amounts of productformed

4. To calculate the amount of product formed ineach case( in moles and in grams)

5. To calculate the amount of reactant in excessleft unreacted)


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How many bicycles can be assembled from the parts shown?

From eightwheels fourbikes can beconstructed.

From fourframes four

bikes can be constructed.

From threepedalassemblies threebikes canbe constructed.

The limitingpart is thenumber ofpedal

assemblies.

Limiting Reactant


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Limiting Reagents

Limiting Reagent is the reactant that is usedup first in a reaction.

It is called the limiting reactant becausethe amount of it present is insufficientto react with the amounts of other

reactants that are present.The limiting reactant limits the

amount of product that can be formed.

Excess reactantis present in quantitiesgreaterthan necessary to react with the

quantity of the limiting reactant


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Limiting Reactant Calculations Method 1

Step 1 Calculate the amount of product (moles or

grams, as needed) formed from each reactant.

Step 2Determine which reactant is limiting. (Thereactant that gives the leastamount of productis the limiting reactant; the other reactant is inexcess.

The limiting reactantis used in all

calculations in a problem


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Excess Reactant

To calculate the amount of the substance that

remains after the reaction ( in excess)

Step 3Calculate the amount of the other reactant

required to react with the limiting reactant, thensubtractthis amount from the starting quantityof the reactant.

Excess mass = given mass mass (reacted)

Excess moles = given moles moles (reacted)


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Limiting Reactant Calculations: Method 2

b

n

a

nBA

aA + bB cC + dDa moles b moles c moles d moles

If A is in excess,

B is the limiting reactant

If

b

n

a

nBA

B is in excess,

A is the limiting reactant

Step 1 Calculate the number of moles of each given

reactant

Step 2Divide the given number of moles of each reactanby the number of moles in the balanced equation


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Theoretical Yieldis the amount of product that wouldresult if allthe limiting reagent reacted.

Actual Yieldis the amount of product actually obtainedfrom a reaction.

% Yield =Actual Yield

Theoretical Yieldx 100

Reaction Yield

The quantities of products calculated from equations

represent the maximum yield(100%) of productaccording to the reaction represented by the

equation.


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SolutionsSolutions

A solution is a homogenous mixture of 2 or more substances

The solute is (are) the substance(s) present in the smalleramount(s)

The solvent is the substance present in the larger amount

The concentrationof a solution is the amount of solute present

in a given quantity of solvent or solution.

M = molarity=moles of solute

Volume of solution (L)

# of moles of solute =

(g/mol)MassMolar

(g)Mass


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1- How many moles are there in 5.0mL of 0.50M

sodium carbonate?

M = n / V , n = M x V

n = 0.50 mol/L x ( 5.0 x 10-3 L)= 2.5 x 10-3mol

2- How many grams of calcium chloride are needed to

prepare 1.0L of 0.50M solution?

M = n / V , n = M x V

n = 0.50 mol/L x 1.0 L = 0.50 mol.

n = mass / molar mass ,

mass = 0.50 mol x 111.1 g/mol = 56 g


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Precipitation Reactions

Precipitate insoluble solid that separates from solution

molecular equation

ionic equation

net ionic equation

Pb2+ + 2NO3- + 2Na+ + 2I- PbI2 (s) + 2Na

+ + 2NO3-

Na+ and NO3- are spectatorionsPbI2

Pb(NO3)2 (aq) + 2NaI (aq) PbI2 (s) + 2NaNO3 (aq)

precipitate

Pb2+ + 2I- PbI2 (s)

A Precipitation reaction is a reaction that occurs in aqueoussolution and results in the formation of an insolubleproduct


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Writing Net Ionic Equations

1. Write the balanced molecular equation.

2. Write the ionic equation showing the strongelectrolytescompletely dissociated into cations and anions.

3. Cancel the spectator ions on both sides of the ionic equation

AgNO3 (aq) + NaCl (aq) AgCl (s) + NaNO3 (aq)

Ag+ + NO3- + Na+ + Cl- AgCl (s) + Na+ + NO3

-

Ag+ + Cl- AgCl (s)

Write the net ionic equation for the reaction of silver

nitrate with sodium chloride.


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Strong electrolytes undergo complete dissociation inwater

KCl (s) K+ (aq) + Cl- (aq)

H2O

1mol 1mol ion 1mol ion

1M KCl solution contains:1 mole of K + ions and 1 mole of Cl- ions

The concentration of the ions: [K+] = 1M; [Cl-] = 1M

[ ] indicate the molar concentrationor Molarity

Ba(NO3)2(s) Ba2+ (aq) + 2NO3

- (aq)

1M Ba(NO3)2 solution [Ba2+] = 1M; [NO3

-] = 2M

H2O


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How many moles of Na+ ions are there in 5.0 ml of

2.0M sodium carbonate (Na2CO3)?

n of sodium carbonate = M x V= 2.0 mol/L x (5.0 x 10-3L) = 0.010 mol

Na2CO3(s) 2 Na+ (aq) + CO3

2-(aq)

n of Na+ = 2 x 0.010 mol = 0.020 mol

H2O


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ExperimentExperiment

Na2CO3(aq) + CaCl2(aq) CaCO3(s) + 2NaCl(aq)

whiteNet ionic equation:

Ca++(aq) + CO32-(aq) CaCO3(s)

Label 5 test tubes of the same diameter. Pipet 10 mL of sodium carbonate(M1) and

10 mL of calcium chloride (M2) in the first

tube. Repeat by varying M1 and M2.

Relate the height of the solid CaCO3 to the

amount of the limiting reagent.


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Report

0.5 M 0.1 M5

0.5 M 1 M4

Reference

X cm

0.5 M 0.5 M3

1 M 0.5 M2

1 M 1 M1

Theoretical

yield of

CaCO3 in

grams

Calculated

Millimoles of

CaCO3formed

Millimoles of

CaCl2 in

10ml of Solution

Used

Millimoles of

Na2CO3 in

10ml of Solution

Used

Comparative

Volume of

Precipitate

Formed*

Using 10ml each

with a

Concentration of:

Na2CO

3CaCl

2

Tube

No.


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5. Decant the solution in the above test tubes and divide thedecantate into two test tubes.

-Add to the first test tube a few ml of 1M calcium chloride-Add to the second test tube a few ml of 1M sodium carbonate.Record your observationsand conclusionsin the table below.

Conclusion

CO32-

Ca2+

Filtrate of

tube 5

Filtrate of

tube 4

Filtrate of

tube 3

Filtrate of

tube 2

Filtrate of

tube 1

Addition of


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6. How many millimolesof the following ions will be left over(if any) after the reaction in each tube?(assuming 100% reaction)

CO32-

Ca2+

Tube 5Tube 4Tube 3Tube 2Tube 1Millimoles

of ions left

101 lab lecture 3 effect of limiting the concentration of

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