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Solution Stoichiometry (Lecture 3)

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

jenette-bailey
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Solution Stoichiometry (Lecture 3)

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

  2. What will I learn? • More examples on Volumetric analysis calculations

  3. 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

  4. 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

  5. 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

  6. 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 

  7. 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 

  8. 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 

  9. 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 

  10. 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 

  11. 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 

  12. 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

  13. 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

  14. 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

  15. 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 

  16. 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 

  17. 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 

  18. 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

  19. 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

  20. 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

  21. 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

  22. 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

  23. 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

  24. 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

  25. 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 

  26. 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 

  27. 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 

  28. 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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