Download presentation

Presentation is loading. Please wait.

Published byJuan Graham Modified over 4 years ago

1
Solution Stoichiometry (Lecture 3) More examples on volumetric analysis calculations

2
YISHUN JC What will I learn? More examples on Volumetric analysis calculations

3
YISHUN JC Balanced equation Example 3 Calculate the volume of nitric acid of concentration 0.200 moldm -3 required to react completely with 4.0 g of copper(II) oxide. v?? Given mass Given c

4
YISHUN JC Balanced equation Example 3 Calculate the volume of nitric acid of concentration 0.200 moldm -3 required to react completely with 4.0 g of copper(II) oxide.

5
YISHUN JC Balanced equation Example 3 Calculate the volume of nitric acid of concentration 0.200 moldm -3 required to react completely with 4.0 g of copper(II) oxide.

6
YISHUN JC Balanced equation Example 3 Calculate the volume of nitric acid of concentration 0.200 moldm -3 required to react completely with 4.0 g of copper(II) oxide.

7
YISHUN JC Balanced equation Example 3 Calculate the volume of nitric acid of concentration 0.200 moldm -3 required to react completely with 4.0 g of copper(II) oxide.

8
YISHUN JC Balanced equation Example 3 Calculate the volume of nitric acid of concentration 0.200 moldm -3 required to react completely with 4.0 g of copper(II) oxide.

9
YISHUN JC Balanced equation Example 3 Calculate the volume of nitric acid of concentration 0.200 moldm -3 required to react completely with 4.0 g of copper(II) oxide.

10
YISHUN JC Balanced equation Example 3 Calculate the volume of nitric acid of concentration 0.200 moldm -3 required to react completely with 4.0 g of copper(II) oxide.

11
YISHUN JC Balanced equation Example 3 Calculate the volume of nitric acid of concentration 0.200 moldm -3 required to react completely with 4.0 g of copper(II) oxide.

12
YISHUN JC Balanced equation Example 4 In an experiment, 25.0 cm 3 of sodium hydroxide of concentration 4.00 gdm -3 completely reacted with 18.0 cm 3 of sulphuric acid. Calculate the molar concentration of the sulphuric acid. c?? Given c (in gdm -3 ), and V Given V

13
YISHUN JC Balanced equation Example 4 In an experiment, 25.0 cm 3 of sodium hydroxide of concentration 4.00 gdm -3 completely reacted with 18.0 cm 3 of sulphuric acid. Calculate the molar concentration of the sulphuric acid.

14
YISHUN JC Balanced equation Example 4 In an experiment, 25.0 cm 3 of sodium hydroxide of concentration 4.00 gdm -3 completely reacted with 18.0 cm 3 of sulphuric acid. Calculate the molar concentration of the sulphuric acid.

15
YISHUN JC Balanced equation Example 4 In an experiment, 25.0 cm 3 of sodium hydroxide of concentration 4.00 gdm -3 completely reacted with 18.0 cm 3 of sulphuric acid. Calculate the molar concentration of the sulphuric acid.

16
YISHUN JC Balanced equation Example 4 In an experiment, 25.0 cm 3 of sodium hydroxide of concentration 4.00 gdm -3 completely reacted with 18.0 cm 3 of sulphuric acid. Calculate the molar concentration of the sulphuric acid.

17
YISHUN JC Balanced equation Example 4 In an experiment, 25.0 cm 3 of sodium hydroxide of concentration 4.00 gdm -3 completely reacted with 18.0 cm 3 of sulphuric acid. Calculate the molar concentration of the sulphuric acid.

18
YISHUN JC Balanced equation Example 5 10.0 g of anhydrous sodium carbonate was dissolved in water, and made up to 250 cm 3 of solution. 25.0 cm 3 of this solution required 20.0 cm 3 of HCl solution for neutralisation. Calculate the concentration of the HCl solution. c?? Given mass in certain V Given V Can calculate c Given V

19
YISHUN JC Balanced equation Example 5 10.0 g of anhydrous sodium carbonate was dissolved in water, and made up to 250 cm 3 of solution. 25.0 cm 3 of this solution required 20.0 cm 3 of HCl solution for neutralisation. Calculate the concentration of the HCl solution.

20
YISHUN JC Balanced equation Example 5 10.0 g of anhydrous sodium carbonate was dissolved in water, and made up to 250 cm 3 of solution. 25.0 cm 3 of this solution required 20.0 cm 3 of HCl solution for neutralisation. Calculate the concentration of the HCl solution.

21
YISHUN JC Balanced equation Example 5 10.0 g of anhydrous sodium carbonate was dissolved in water, and made up to 250 cm 3 of solution. 25.0 cm 3 of this solution required 20.0 cm 3 of HCl solution for neutralisation. Calculate the concentration of the HCl solution.

22
YISHUN JC Balanced equation Example 5 10.0 g of anhydrous sodium carbonate was dissolved in water, and made up to 250 cm 3 of solution. 25.0 cm 3 of this solution required 20.0 cm 3 of HCl solution for neutralisation. Calculate the concentration of the HCl solution.

23
YISHUN JC Balanced equation Example 5 10.0 g of anhydrous sodium carbonate was dissolved in water, and made up to 250 cm 3 of solution. 25.0 cm 3 of this solution required 20.0 cm 3 of HCl solution for neutralisation. Calculate the concentration of the HCl solution.

24
YISHUN JC Balanced equation Example 5 10.0 g of anhydrous sodium carbonate was dissolved in water, and made up to 250 cm 3 of solution. 25.0 cm 3 of this solution required 20.0 cm 3 of HCl solution for neutralisation. Calculate the concentration of the HCl solution.

25
YISHUN JC Balanced equation Example 5 10.0 g of anhydrous sodium carbonate was dissolved in water, and made up to 250 cm 3 of solution. 25.0 cm 3 of this solution required 20.0 cm 3 of HCl solution for neutralisation. Calculate the concentration of the HCl solution.

26
YISHUN JC Balanced equation Example 5 10.0 g of anhydrous sodium carbonate was dissolved in water, and made up to 250 cm 3 of solution. 25.0 cm 3 of this solution required 20.0 cm 3 of HCl solution for neutralisation. Calculate the concentration of the HCl solution.

27
YISHUN JC Balanced equation Example 5 10.0 g of anhydrous sodium carbonate was dissolved in water, and made up to 250 cm 3 of solution. 25.0 cm 3 of this solution required 20.0 cm 3 of HCl solution for neutralisation. Calculate the concentration of the HCl solution.

28
YISHUN JC What have I learnt? More examples on Volumetric analysis calculations

29
End of Lecture 3 Often greater risk is involved in postponement than in making a wrong decision Harry A Hopf

Similar presentations

OK

Learning objective: To calculate expected and percentage yield 09/06/2016 On whiteboards…

Learning objective: To calculate expected and percentage yield 09/06/2016 On whiteboards…

© 2018 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on different aspects of environment and society Ppt on column chromatography lab Ppt on forward rate agreement example Ppt on artificial intelligence and neural networks Ppt on metro rail project Ppt on national education policy 1986 calendar Gastrointestinal anatomy and physiology ppt on cells Ppt on negotiation skills Ppt on power quality and control Ppt on travelling salesman problem using genetic algorithm