lecture thirteen chm 151 ©slg topics: 1. molarity 1. solution stoichiometry 2. limiting reagent...

LECTURE ThirteenCHM 151 ©slg

Topics:

1. Molarity1. Solution Stoichiometry 2. Limiting Reagent Solution Problems

Net Ionic Equations: What to “Break Up”:

A) all soluble salts “(aq)”

B) All soluble bases “(aq)”

C) All strong acids: HCl HBr HI; HNO3 H2SO4 HClO4

What not to “break up”:

A) all insoluble salts or bases “(s)”

B) H2O (and all molecules)

C) All weak acids: H3PO4, HCH3CO2, H2CO3.....

H3PO4 (aq) + KOH (aq) ----> ? -----> H2O + salt

Step One: salt formula

K+ + PO43- -----> K3PO4 (aq)

Step Two: Write Equation; Balance

H3PO4 (aq) + KOH (aq) ----> H2O (l) + K3PO4 (aq)

H3PO4 (aq) + 3 KOH (aq) ---->3 H2O (l) + K3PO4 (aq)

H3PO4 (aq) + 3 KOH (aq) ---->3 H2O (l) + K3PO4 (aq)

Step Three: Total Ionic:

H3PO4 (aq) + 3 K+ (aq) + 3 OH- (aq) -----> 3 H2O (l)

+ 3 K+ (aq) + PO43-(aq)

Step Four: Net Ionic:

H3PO4 (aq) + 3 OH- (aq) -----> 3 H2O (l) + PO43-(aq)

Weak acid, no! Yes! No! Yes!

HCH3CO2 (aq) + Mg(OH)2 (s) ----->? -----> H2O + salt

Step One: Salt Formula:

Mg2+ + CH3CO2- -----> Mg(CH3CO2)2 (aq)

Step Two: Write Equation, Balance:

HCH3CO2 (aq) + Mg(OH)2(s) -----> H2O + Mg(CH3CO2)2(aq)

2 HCH3CO2 (aq) + Mg(OH)2(s) -----> 2 H2O + Mg(CH3CO2)2(aq)

2 HCH3CO2 (aq) + Mg(OH)2(s) -----> 2 H2O + Mg(CH3CO2)2(aq)

Step Three: Total Ionic Equation:

2 HCH3CO2 (aq) + Mg(OH)2(s) -----> 2 H2O(l) + Mg2+ (aq)

+2 CH3CO2- (aq)

Step Four: Net Ionic Equation

SAME!

NO, weak acid! NO! NO! Yes!

Concentrations of Compounds in Aqueous Solutions

(Chapter 5, Section 5.8, p. 213)

Usually the reactions we run are done in aqueous solution, and therefore we need to add to our study of stoichiometry the concentration of compounds in aqueous solution.

We will utilize a very useful quantity known as “molarity,” the number of moles of solute per liter of solution.

Concentration(molarity) = # moles solute L solution

If we placed 1.00 mol NaCl (58.4 g) in a 1 L volumetric flask, dissolved it in water, swirled to dissolve and diluted the solution to the 1.00 liter mark you would have a solution that contains 1.00 mol NaCl per liter of solution. This may be represented several ways:

Concentration(molarity) = 1.00 mol NaCl / L soln = 1.00 M NaCl = [1.00] NaCl

Chemists call this a “1.00 molar solution”

Calculating Molar Amounts

TYPICAL PROBLEMS:

•What is the molarity of a solution made by dissolving25.0 g of BaCl2 in sufficient water to make up a solution of 500.0 mL?

•How many g of BaCl2 would be contained in 20.0 mL ofthis solution?

•How many mL of this solution would deliver 1.25 g ofBaCl2?

•How many mol of Cl- ions are contained in 10.00 ml ofthis solution?

What is the molarity of a solution made by dissolving25.0 g of BaCl2 in sufficient water to make up a solution of 500.0 mL?

25.0 g BaCl2 = ? mol/ L BaCl2 (= ? M BaCl2) 500.0 mL soln

1Ba = 1 X 137.33g = 137.332Cl = 2 X 35.45g = 70.90 208.23 g/mol

Molar mass, BaCl2:

25.0 g BaCl2

208.23 g BaCl2

1 mol BaCl2

500.0 mL

1000 mL

1 L= .240 mol / L BaCl2 = .240 M BaCl2

Mass of solute

Volume of soltn

Molar mass,solute

Conversionto L

How many g of BaCl2 would be contained in 20.0 mL ofthis solution? (.240 M BaCl2)

Question: 20.0 mL soln = ? g BaCl2

Relationships: 1000 mL = 1 L 1 L soln = .240 mol BaCl2

1 mol BaCl2 = 208.23 g BaCl2

20.0 mL soln = ? g BaCl2

mL L mol g

20.0 mL soln

1000 mL

1 L

1L soln

.240 mol BaCl2

1 mol BaCl2

208.23 g BaCl2 = 1.00 g BaCl2

Molarity Molar Mass

conversion

•How many mL of this solution would deliver 1.25 g ofBaCl2?

1.25 g BaCl2 = ? mL soln.240 mol BaCl2 = 1 L soln208.23 g BaCl2 = 1 mol BaCl2

1.25 g BaCl2

208.23 g BaCl2

1 mol BaCl2

.240 mol BaCl2

1L soln

1 L soln

1000 mL= 25.0 mL

Molar Mass Molarity

•How many mol of Cl- ions are contained in 10.0 ml ofthis solution?

10.0 mL soln = ? Mol Cl- .240 mol BaCl2 = 1000 mL soln 1 mol BaCl2 = 2 mol Cl-

Note: BaCl2(aq) ---> Ba2+(aq) + 2 Cl- (aq)

10.0 mL soln

1000 mL soln

.240 mol BaCl2

1 mol BaCl2

2 mol Cl-

= .00480 mol Cl-

GROUP WORK:

If 35.00 g CuSO4 is dissolved in sufficient water to makeup 750. mL of an aqueous solution,

a) what is the molarity of the solution?

b) how many mL of the solution will deliver 10.0 g of CuSO4?

c) How many moles of sulfate ion (SO42-) will be

delivered in 10.0 mL of the solution?

1 Cu = 1 X 63.55 = 63.551 S = 1 X 32.07 = 32.074 O = 4 X 16.00 = 64.00 159.62 g/mol

35.00 g CuSO4

750 mL 159.62 g CuSO4

1 mol CuSO 4 1000 mL

1 L= .292 mol/L CuSO4

= .292 M CuSO4

10.0 g CuSO4

159.62 g CuSO4

1 mol CuSO 4

.292 mol CuSO4

1000 mL soln= 215 mL soln

10.0 mL soln

1000 mL soln

.292 mol CuSO4

1 mol CuSO4

1 mol SO42-

= .00292 mol SO42-

STOICHIOMETRY OF REACTIONS IN AQUEOUSSOLUTION: Chapter 5, Section 5.9

Let’s use our favorite reaction to add another dimension to calculating from balanced equations:

How many ml of 3.00 M HCl solution would be requiredto react with 13.67 g of Fe2O3 according to the followingbalanced equation:

Fe2O3(s) + 6HCl(aq) ---> 3H2O + 2FeCl3(aq)

159.70 g/mol 36.46 g/mol 18.02 g/mol 162.20 g/mol 3.00 M13.67 g ? mL


159.70 g/mol 3.00 M 13.67 g ? mL

Pathway: g Fe2O3 ---> mol Fe2O3---> mol HCl --- > mL soln

13.67 g Fe2O3 = ? mL soln

1000 mL soln = 3.00 mol HCl 6 mol HCl = 1 mol Fe2O3

159.70 g Fe2O3 = 1 mol Fe2O3

13.67 g Fe2O3 = ? mL soltn

13.67 g Fe2O3

159.70 gFe2O3

1 mol Fe2O3

1 mol Fe2O3

6 mol HCl

3.00 mol HCl

1000 mL soln= 171 mL soln

MolarMass

BalancedEquation

Molarity

Group Work

How many g of Fe2O3 would react with 25.0 mL of 3.00 M HCl?


159.70 g/mol 3.00 M ? g 25.0 mL

mL mol HCl mol Fe2O3 g Fe2O3

25.0 mL soltn = ? g Fe2O3


159.70 g/mol 3.00 M ? g 25.0 mL

25.0 mL soln 3.00 mol HCl 1 mol Fe2O3 159.70 g Fe2O3

1000 mL soln 6 mol HCl 1 mol Fe2O3

= 1.996 g Fe2O3 = 2.00 g Fe2O3

Molarity EquationMolar mass

Limiting Reagent, Solutions

Suppose you mixed 20.00 mL of .250 M Pb(NO3)2 solutionwith 30.00 mL of .150 KI solution. How many g of PbI2

precipitate might you theoretically obtain?

Pb(NO3)2 (aq) + 2 KI (aq) ---> PbI2 (s) + 2 KNO3 (aq)

.250 M .150 M 461.0 g/mol

20.00 mL 30.00 mL ? g1Pb = 1 X 207.2 = 207.22 I = 2 X 126.9 = 253.8 461.0 g/mol


.250 M .150 M 461.0 g/mol20.00 mL 30.00 mL ? g

Pathway: mL mol “A” mol “C” g “C”

mL mol “B” mol “C” g “C”

“A” “B” “C”

Lownumberwins!


.250 M .150 M 461.0 g/mol20.00 mL 30.00 mL ? g

20.00 ml soltn A .250 mol A 1 mol C 461.0 g C = 2.3050 g C 1000 mL 1 mol A 1 mol C = 2.30 g C

30.00 mL soltn B .150 mol B 1 mol C 461.0 g C = 1.037 g C 1000 mL 2 mol B 1 mol C

= 1.04 g C

How many mL of solution A, .250 M Pb(NO3)2, would be required to react with 30.00 mL of solution B, .150 M KI?


.250 M .150 M ? mL 30.00 mL

Excess reagent:

Pathway: mL “B” soltn mol “B” mol “A” mL “A” soltn

30.00 mL B soltn = ? mL A soltn


.250 M .150 M ? mL 30.00 mL

30.00 mL soltn .150 mol B 1 mol A 1000 mL soltn 1000 mL soltn 2 mol B .250 mol A

= 9.00 mL soltn A

Molarity equation Molarity

Proof:

9.00 mL soltn A .250 Mol A 2 mol B = .00450 mol B 1000 mL soltn 1 mol A


.250 M .150 M 9.00 mL 30.00 mL

30.00 mL soltn B .150 mol B = .00450 mol B 1000 mL soltn

Group Work:

How many mL of .150 M KI solution would be required to react with 20.00 mL of .250 M Pb(NO3)2 ?


.250 M .150 M 20.00 mL ? mL

Answer:


.250 M .150 M 20.00 mL ? mL

20.00 mL soltn A .250 mol A 2 mol B 1000 mL soltn B 1000 mL soltn A 1 mol A .150 mol B

= 66.67 mL soltn B

Summary

1. Molarity, M: useful description of a solution; givesnumber of moles of solute in 1 L of solution. Usefulto convert mL of solution to moles of solute in equationsituations.

2. Molarity of solution can also yield moles per liter ofthe ions produced when the solute ionizes in water,using the formula of the solute as conversion factor

3. Adds another way to calculate mass or moles from volume (compare to density, #grams of substanceor solution in 1 mL or 1 cm3 of solution or substance).

lecture thirteen chm 151 ©slg topics: 1. molarity 1. solution stoichiometry 2. limiting reagent...

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