solution stoichiometry lecture 3
Download
Skip this Video
Download Presentation
Solution Stoichiometry (Lecture 3)

Loading in 2 Seconds...

play fullscreen
1 / 29

Solution Stoichiometry (Lecture 3) - PowerPoint PPT Presentation


  • 93 Views
  • Uploaded on

Solution Stoichiometry (Lecture 3). More examples on volumetric analysis calculations. What will I learn?. More examples on Volumetric analysis calculations.

loader
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
capcha
Download Presentation

PowerPoint Slideshow about 'Solution Stoichiometry (Lecture 3)' - jenette-bailey


An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
solution stoichiometry lecture 3

Solution Stoichiometry(Lecture 3)

More examples on volumetric analysis calculations

what will i learn
What will I learn?
  • More examples on Volumetric analysis calculations
slide3
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.

Balanced equation

Given mass

v??

Given c

slide4
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.

Balanced equation

slide5
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.

Balanced equation

slide6
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.

Balanced equation

slide7
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.

Balanced equation

slide8
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.

Balanced equation

slide9
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.

Balanced equation

slide10
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.

Balanced equation

slide11
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.

Balanced equation

slide12
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.

Balanced equation

Given c (in gdm-3), and V

c??

Given V

slide13
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.

Balanced equation

slide14
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.

Balanced equation

slide15
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.

Balanced equation

slide16
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.

Balanced equation

slide17
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.

Balanced equation

slide18
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.

Balanced equation

c??

Given mass in certain V

Given V

 Can calculate c

Given V

slide19
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.

Balanced equation

slide20
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.

Balanced equation

slide21
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.

Balanced equation

slide22
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.

Balanced equation

slide23
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.

Balanced equation

slide24
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.

Balanced equation

slide25
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.

Balanced equation

slide26
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.

Balanced equation

slide27
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.

Balanced equation

what have i learnt
What have I learnt?
  • More examples on Volumetric analysis calculations
end of lecture 3

End of Lecture 3

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

ad