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

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Solution Stoichiometry (Lecture 3)

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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


Solution stoichiometry lecture 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


Solution stoichiometry lecture 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


Solution stoichiometry lecture 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


Solution stoichiometry lecture 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


Solution stoichiometry lecture 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


Solution stoichiometry lecture 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


Solution stoichiometry lecture 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


Solution stoichiometry lecture 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


Solution stoichiometry lecture 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


Solution stoichiometry lecture 3

  • 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


Solution stoichiometry lecture 3

  • 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


Solution stoichiometry lecture 3

  • 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


Solution stoichiometry lecture 3

  • 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


Solution stoichiometry lecture 3

  • 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


Solution stoichiometry lecture 3

  • 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


Solution stoichiometry lecture 3

  • 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


Solution stoichiometry lecture 3

  • 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


Solution stoichiometry lecture 3

  • 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


Solution stoichiometry lecture 3

  • 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


Solution stoichiometry lecture 3

  • 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


Solution stoichiometry lecture 3

  • 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


Solution stoichiometry lecture 3

  • 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


Solution stoichiometry lecture 3

  • 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


Solution stoichiometry lecture 3

  • 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


Solution stoichiometry lecture 3

  • 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


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