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

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Presentation on theme: "Solution Stoichiometry (Lecture 3) More examples on volumetric analysis calculations."— Presentation transcript:

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 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 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 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 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 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 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 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 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 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 g of anhydrous sodium carbonate was dissolved in water, and made up to 250 cm 3 of solution 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 g of anhydrous sodium carbonate was dissolved in water, and made up to 250 cm 3 of solution 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 g of anhydrous sodium carbonate was dissolved in water, and made up to 250 cm 3 of solution 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 g of anhydrous sodium carbonate was dissolved in water, and made up to 250 cm 3 of solution 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 g of anhydrous sodium carbonate was dissolved in water, and made up to 250 cm 3 of solution 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 g of anhydrous sodium carbonate was dissolved in water, and made up to 250 cm 3 of solution 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 g of anhydrous sodium carbonate was dissolved in water, and made up to 250 cm 3 of solution 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 g of anhydrous sodium carbonate was dissolved in water, and made up to 250 cm 3 of solution 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 g of anhydrous sodium carbonate was dissolved in water, and made up to 250 cm 3 of solution 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 g of anhydrous sodium carbonate was dissolved in water, and made up to 250 cm 3 of solution 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


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