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

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Solution Stoichiometry (Lecture 3) More examples on volumetric analysis calculations Slide 2 YISHUN JC What will I learn? More examples on Volumetric analysis calculations Slide 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 Slide 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. Slide 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. Slide 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. Slide 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. Slide 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. Slide 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. Slide 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. Slide 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. Slide 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 Slide 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. Slide 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. Slide 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. Slide 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. Slide 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. Slide 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 Slide 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. Slide 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. Slide 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. Slide 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. Slide 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. Slide 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. Slide 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. Slide 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. Slide 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. Slide 28 YISHUN JC What have I learnt? More examples on Volumetric analysis calculations Slide 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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