Solution Stoichiometry(Lecture 3)

More examples on volumetric analysis calculations

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What will I learn? More examples on Volumetric analysis

calculations

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Balanced equation

Example 3Calculate the volume of nitric acid of concentration 0.200 moldm-3 required to react completely with 4.0 g of copper(II) oxide.

OHNOCuCuOHNO 22332

v?? Given massGiven c

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Balanced equation

Example 3Calculate the volume of nitric acid of concentration 0.200 moldm-3 required to react completely with 4.0 g of copper(II) oxide.

OHNOCuCuOHNO 22332

molgmol

g

M

mn

CuO

CuOCuO 0503.0

0.165.63

0.41

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Balanced equation

Example 3Calculate the volume of nitric acid of concentration 0.200 moldm-3 required to react completely with 4.0 g of copper(II) oxide.

OHNOCuCuOHNO 22332

molgmol

g

M

mn

CuO

CuOCuO 0503.0

0.165.63

0.41

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Balanced equation

Example 3Calculate the volume of nitric acid of concentration 0.200 moldm-3 required to react completely with 4.0 g of copper(II) oxide.

OHNOCuCuOHNO 22332

molgmol

g

M

mn

CuO

CuOCuO 0503.0

0.165.63

0.41

1

23

CuO

HNO

n

nmolnn CuOHNO 101.00503.02

1

23

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Balanced equation

Example 3Calculate the volume of nitric acid of concentration 0.200 moldm-3 required to react completely with 4.0 g of copper(II) oxide.

OHNOCuCuOHNO 22332

molgmol

g

M

mn

CuO

CuOCuO 0503.0

0.165.63

0.41

1

23

CuO

HNO

n

nmolnn CuOHNO 101.00503.02

1

23

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Balanced equation

Example 3Calculate the volume of nitric acid of concentration 0.200 moldm-3 required to react completely with 4.0 g of copper(II) oxide.

OHNOCuCuOHNO 22332

molgmol

g

M

mn

CuO

CuOCuO 0503.0

0.165.63

0.41

1

23

CuO

HNO

n

nmolnn CuOHNO 101.00503.02

1

23

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Balanced equation

Example 3Calculate the volume of nitric acid of concentration 0.200 moldm-3 required to react completely with 4.0 g of copper(II) oxide.

OHNOCuCuOHNO 22332

molgmol

g

M

mn

CuO

CuOCuO 0503.0

0.165.63

0.41

1

23

CuO

HNO

n

nmolnn CuOHNO 101.00503.02

1

23

33

503.0200.0

101.0

3

3

3dm

moldm

mol

c

nV

HNO

HNOHNO

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Balanced equation

Example 3Calculate the volume of nitric acid of concentration 0.200 moldm-3 required to react completely with 4.0 g of copper(II) oxide.

OHNOCuCuOHNO 22332

molgmol

g

M

mn

CuO

CuOCuO 0503.0

0.165.63

0.41

1

23

CuO

HNO

n

nmolnn CuOHNO 101.00503.02

1

23

33

503.0200.0

101.0

3

3

3dm

moldm

mol

c

nV

HNO

HNOHNO

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Balanced equation

Example 3Calculate the volume of nitric acid of concentration 0.200 moldm-3 required to react completely with 4.0 g of copper(II) oxide.

OHNOCuCuOHNO 22332

molgmol

g

M

mn

CuO

CuOCuO 0503.0

0.165.63

0.41

1

23

CuO

HNO

n

nmolnn CuOHNO 101.00503.02

1

23

33

503.0200.0

101.0

3

3

3dm

moldm

mol

c

nV

HNO

HNOHNO

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Balanced equation

Example 4In an experiment, 25.0 cm3 of sodium hydroxide of concentration 4.00 gdm-3 completely reacted with 18.0 cm3 of sulphuric acid. Calculate the molar concentration of the sulphuric acid.

OHSONaSOHNaOH 24242 22

c??Given c (in gdm-3),

and V Given V

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)

))

1-

-33-

(gmol mass molar

(gdm NaOH of ionconcentrat(moldm NaOH of ionconcentrat

1-

-3

1.0)gmol16.0(23.0

gdm 4.00

Balanced equation

OHSONaSOHNaOH 24242 22

Example 4In an experiment, 25.0 cm3 of sodium hydroxide of concentration 4.00 gdm-3 completely reacted with 18.0 cm3 of sulphuric acid. Calculate the molar concentration of the sulphuric acid.

31

3

100.00.40

0.4

moldmgmol

gdm

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)

))

1-

-33-

(gmol mass molar

(gdm NaOH of ionconcentrat(moldm NaOH of ionconcentrat

1-

-3

1.0)gmol16.0(23.0

gdm 4.00

Balanced equation

OHSONaSOHNaOH 24242 22

Example 4In an experiment, 25.0 cm3 of sodium hydroxide of concentration 4.00 gdm-3 completely reacted with 18.0 cm3 of sulphuric acid. Calculate the molar concentration of the sulphuric acid.

31

3

100.00.40

0.4

moldmgmol

gdm

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)

))

1-

-33-

(gmol mass molar

(gdm NaOH of ionconcentrat(moldm NaOH of ionconcentrat

Balanced equation

OHSONaSOHNaOH 24242 22

Example 4In an experiment, 25.0 cm3 of sodium hydroxide of concentration 4.00 gdm-3 completely reacted with 18.0 cm3 of sulphuric acid. Calculate the molar concentration of the sulphuric acid.

3100.0 moldm

2

142

NaOH

SOH

n

n

2

14242

NaOHNaOH

SOHSOH

Vc

Vc

33

3

0694.0100.18

100.25100.0

2

1

2

1

42

42

moldm

V

Vcc

SOH

NaOHNaOHSOH

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)

))

1-

-33-

(gmol mass molar

(gdm NaOH of ionconcentrat(moldm NaOH of ionconcentrat

Balanced equation

OHSONaSOHNaOH 24242 22

Example 4In an experiment, 25.0 cm3 of sodium hydroxide of concentration 4.00 gdm-3 completely reacted with 18.0 cm3 of sulphuric acid. Calculate the molar concentration of the sulphuric acid.

3100.0 moldm

2

142

NaOH

SOH

n

n

2

14242

NaOHNaOH

SOHSOH

Vc

Vc

33

3

0694.0100.18

100.25100.0

2

1

2

1

42

42

moldm

V

Vcc

SOH

NaOHNaOHSOH

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)

))

1-

-33-

(gmol mass molar

(gdm NaOH of ionconcentrat(moldm NaOH of ionconcentrat

Balanced equation

OHSONaSOHNaOH 24242 22

Example 4In an experiment, 25.0 cm3 of sodium hydroxide of concentration 4.00 gdm-3 completely reacted with 18.0 cm3 of sulphuric acid. Calculate the molar concentration of the sulphuric acid.

3100.0 moldm

2

142

NaOH

SOH

n

n

2

14242

NaOHNaOH

SOHSOH

Vc

Vc

33

3

0694.0100.18

100.25100.0

2

1

2

1

42

42

moldm

V

Vcc

SOH

NaOHNaOHSOH

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Balanced equation

Example 510.0 g of anhydrous sodium carbonate was dissolved in water, and made up to 250 cm3 of solution. 25.0 cm3 of this solution required 20.0 cm3 of HCl solution for neutralisation. Calculate the concentration of the HCl solution.

c??Given mass in certain V

Given V

OHCONaClHClCONa 2232 22

Can calculate c

Given V

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Balanced equation

Example 510.0 g of anhydrous sodium carbonate was dissolved in water, and made up to 250 cm3 of solution. 25.0 cm3 of this solution required 20.0 cm3 of HCl solution for neutralisation. Calculate the concentration of the HCl solution.

OHCONaClHClCONa 2232 22

11 0.106

0.10

0.1630.120.232

0.10

32

32

32

gmol

g

gmol

g

M

mn

CONa

CONaCONa

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Balanced equation

Example 510.0 g of anhydrous sodium carbonate was dissolved in water, and made up to 250 cm3 of solution. 25.0 cm3 of this solution required 20.0 cm3 of HCl solution for neutralisation. Calculate the concentration of the HCl solution.

OHCONaClHClCONa 2232 22

11 0.106

0.10

0.1630.120.232

0.10

32

32

32

gmol

g

gmol

g

M

mn

CONa

CONaCONa

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Balanced equation

Example 510.0 g of anhydrous sodium carbonate was dissolved in water, and made up to 250 cm3 of solution. 25.0 cm3 of this solution required 20.0 cm3 of HCl solution for neutralisation. Calculate the concentration of the HCl solution.

OHCONaClHClCONa 2232 22

11 0.106

0.10

0.1630.120.232

0.10

32

32

32

gmol

g

gmol

g

M

mn

CONa

CONaCONa

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Balanced equation

Example 510.0 g of anhydrous sodium carbonate was dissolved in water, and made up to 250 cm3 of solution. 25.0 cm3 of this solution required 20.0 cm3 of HCl solution for neutralisation. Calculate the concentration of the HCl solution.

OHCONaClHClCONa 2232 22

11 0.106

0.10

0.1630.120.232

0.10

32

32

32

gmol

g

gmol

g

M

mn

CONa

CONaCONa mol0943.0

333

377.010250

0943.032

32

moldm

dm

mol

V

nc CONa

CONa

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Balanced equation

Example 510.0 g of anhydrous sodium carbonate was dissolved in water, and made up to 250 cm3 of solution. 25.0 cm3 of this solution required 20.0 cm3 of HCl solution for neutralisation. Calculate the concentration of the HCl solution.

OHCONaClHClCONa 2232 22

11 0.106

0.10

0.1630.120.232

0.10

32

32

32

gmol

g

gmol

g

M

mn

CONa

CONaCONa mol0943.0

333

377.010250

0943.032

32

moldm

dm

mol

V

nc CONa

CONa

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Balanced equation

Example 510.0 g of anhydrous sodium carbonate was dissolved in water, and made up to 250 cm3 of solution. 25.0 cm3 of this solution required 20.0 cm3 of HCl solution for neutralisation. Calculate the concentration of the HCl solution.

OHCONaClHClCONa 2232 22

11 0.106

0.10

0.1630.120.232

0.10

32

32

32

gmol

g

gmol

g

M

mn

CONa

CONaCONa mol0943.0

333

377.010250

0943.032

32

moldm

dm

mol

V

nc CONa

CONa

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Balanced equation

Example 510.0 g of anhydrous sodium carbonate was dissolved in water, and made up to 250 cm3 of solution. 25.0 cm3 of this solution required 20.0 cm3 of HCl solution for neutralisation. Calculate the concentration of the HCl solution.

OHCONaClHClCONa 2232 22

333

377.010250

0943.032

32

moldm

dm

mol

V

nc CONa

CONa

1

2

32

CONa

HCl

n

n 1

2

3232

CONaCONa

HClHCl

Vc

Vc

33

3

943.0100.20

100.25377.0

1

2

1

23232

moldm

V

Vcc

HCl

CONaCONaHCl

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Balanced equation

Example 510.0 g of anhydrous sodium carbonate was dissolved in water, and made up to 250 cm3 of solution. 25.0 cm3 of this solution required 20.0 cm3 of HCl solution for neutralisation. Calculate the concentration of the HCl solution.

OHCONaClHClCONa 2232 22

333

377.010250

0943.032

32

moldm

dm

mol

V

nc CONa

CONa

1

2

32

CONa

HCl

n

n 1

2

3232

CONaCONa

HClHCl

Vc

Vc

33

3

943.0100.20

100.25377.0

1

2

1

23232

moldm

V

Vcc

HCl

CONaCONaHCl

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Balanced equation

Example 510.0 g of anhydrous sodium carbonate was dissolved in water, and made up to 250 cm3 of solution. 25.0 cm3 of this solution required 20.0 cm3 of HCl solution for neutralisation. Calculate the concentration of the HCl solution.

OHCONaClHClCONa 2232 22

333

377.010250

0943.032

32

moldm

dm

mol

V

nc CONa

CONa

1

2

32

CONa

HCl

n

n 1

2

3232

CONaCONa

HClHCl

Vc

Vc

33

3

943.0100.20

100.25377.0

1

2

1

23232

moldm

V

Vcc

HCl

CONaCONaHCl

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What have I learnt? More examples on Volumetric analysis

calculations

End of Lecture 3

“Often greater risk is involved in postponement than in making a wrong decision” Harry A Hopf