solution stoichiometry (lecture 3)
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Solution Stoichiometry (Lecture 3). More examples on volumetric analysis calculations. What will I learn?. More examples on Volumetric analysis calculations. - PowerPoint PPT PresentationTRANSCRIPT
![Page 1: Solution Stoichiometry (Lecture 3)](https://reader036.vdocuments.mx/reader036/viewer/2022062321/56813655550346895d9ddc4e/html5/thumbnails/1.jpg)
Solution Stoichiometry(Lecture 3)
More examples on volumetric analysis calculations
![Page 2: Solution Stoichiometry (Lecture 3)](https://reader036.vdocuments.mx/reader036/viewer/2022062321/56813655550346895d9ddc4e/html5/thumbnails/2.jpg)
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What will I learn? More examples on Volumetric analysis
calculations
![Page 3: Solution Stoichiometry (Lecture 3)](https://reader036.vdocuments.mx/reader036/viewer/2022062321/56813655550346895d9ddc4e/html5/thumbnails/3.jpg)
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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
![Page 4: Solution Stoichiometry (Lecture 3)](https://reader036.vdocuments.mx/reader036/viewer/2022062321/56813655550346895d9ddc4e/html5/thumbnails/4.jpg)
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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
![Page 5: Solution Stoichiometry (Lecture 3)](https://reader036.vdocuments.mx/reader036/viewer/2022062321/56813655550346895d9ddc4e/html5/thumbnails/5.jpg)
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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
![Page 6: Solution Stoichiometry (Lecture 3)](https://reader036.vdocuments.mx/reader036/viewer/2022062321/56813655550346895d9ddc4e/html5/thumbnails/6.jpg)
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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
![Page 7: Solution Stoichiometry (Lecture 3)](https://reader036.vdocuments.mx/reader036/viewer/2022062321/56813655550346895d9ddc4e/html5/thumbnails/7.jpg)
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HU
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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
![Page 8: Solution Stoichiometry (Lecture 3)](https://reader036.vdocuments.mx/reader036/viewer/2022062321/56813655550346895d9ddc4e/html5/thumbnails/8.jpg)
YIS
HU
N J
C
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
![Page 9: Solution Stoichiometry (Lecture 3)](https://reader036.vdocuments.mx/reader036/viewer/2022062321/56813655550346895d9ddc4e/html5/thumbnails/9.jpg)
YIS
HU
N J
C
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
![Page 10: Solution Stoichiometry (Lecture 3)](https://reader036.vdocuments.mx/reader036/viewer/2022062321/56813655550346895d9ddc4e/html5/thumbnails/10.jpg)
YIS
HU
N J
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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
![Page 11: Solution Stoichiometry (Lecture 3)](https://reader036.vdocuments.mx/reader036/viewer/2022062321/56813655550346895d9ddc4e/html5/thumbnails/11.jpg)
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N J
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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
![Page 12: Solution Stoichiometry (Lecture 3)](https://reader036.vdocuments.mx/reader036/viewer/2022062321/56813655550346895d9ddc4e/html5/thumbnails/12.jpg)
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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
![Page 13: Solution Stoichiometry (Lecture 3)](https://reader036.vdocuments.mx/reader036/viewer/2022062321/56813655550346895d9ddc4e/html5/thumbnails/13.jpg)
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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
![Page 14: Solution Stoichiometry (Lecture 3)](https://reader036.vdocuments.mx/reader036/viewer/2022062321/56813655550346895d9ddc4e/html5/thumbnails/14.jpg)
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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
![Page 15: Solution Stoichiometry (Lecture 3)](https://reader036.vdocuments.mx/reader036/viewer/2022062321/56813655550346895d9ddc4e/html5/thumbnails/15.jpg)
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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
![Page 16: Solution Stoichiometry (Lecture 3)](https://reader036.vdocuments.mx/reader036/viewer/2022062321/56813655550346895d9ddc4e/html5/thumbnails/16.jpg)
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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
![Page 17: Solution Stoichiometry (Lecture 3)](https://reader036.vdocuments.mx/reader036/viewer/2022062321/56813655550346895d9ddc4e/html5/thumbnails/17.jpg)
YIS
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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
![Page 18: Solution Stoichiometry (Lecture 3)](https://reader036.vdocuments.mx/reader036/viewer/2022062321/56813655550346895d9ddc4e/html5/thumbnails/18.jpg)
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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
![Page 19: Solution Stoichiometry (Lecture 3)](https://reader036.vdocuments.mx/reader036/viewer/2022062321/56813655550346895d9ddc4e/html5/thumbnails/19.jpg)
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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
![Page 20: Solution Stoichiometry (Lecture 3)](https://reader036.vdocuments.mx/reader036/viewer/2022062321/56813655550346895d9ddc4e/html5/thumbnails/20.jpg)
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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
![Page 21: Solution Stoichiometry (Lecture 3)](https://reader036.vdocuments.mx/reader036/viewer/2022062321/56813655550346895d9ddc4e/html5/thumbnails/21.jpg)
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HU
N J
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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
![Page 22: Solution Stoichiometry (Lecture 3)](https://reader036.vdocuments.mx/reader036/viewer/2022062321/56813655550346895d9ddc4e/html5/thumbnails/22.jpg)
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N J
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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
![Page 23: Solution Stoichiometry (Lecture 3)](https://reader036.vdocuments.mx/reader036/viewer/2022062321/56813655550346895d9ddc4e/html5/thumbnails/23.jpg)
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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
![Page 24: Solution Stoichiometry (Lecture 3)](https://reader036.vdocuments.mx/reader036/viewer/2022062321/56813655550346895d9ddc4e/html5/thumbnails/24.jpg)
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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
![Page 25: Solution Stoichiometry (Lecture 3)](https://reader036.vdocuments.mx/reader036/viewer/2022062321/56813655550346895d9ddc4e/html5/thumbnails/25.jpg)
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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
![Page 26: Solution Stoichiometry (Lecture 3)](https://reader036.vdocuments.mx/reader036/viewer/2022062321/56813655550346895d9ddc4e/html5/thumbnails/26.jpg)
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N J
C
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
![Page 27: Solution Stoichiometry (Lecture 3)](https://reader036.vdocuments.mx/reader036/viewer/2022062321/56813655550346895d9ddc4e/html5/thumbnails/27.jpg)
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N J
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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
![Page 28: Solution Stoichiometry (Lecture 3)](https://reader036.vdocuments.mx/reader036/viewer/2022062321/56813655550346895d9ddc4e/html5/thumbnails/28.jpg)
YIS
HU
N J
C
What have I learnt? More examples on Volumetric analysis
calculations
![Page 29: Solution Stoichiometry (Lecture 3)](https://reader036.vdocuments.mx/reader036/viewer/2022062321/56813655550346895d9ddc4e/html5/thumbnails/29.jpg)
End of Lecture 3
“Often greater risk is involved in postponement than in making a wrong decision” Harry A Hopf