Stoichiometry In Solution Chemistry. Stochiometry involves calculating the amounts of reactants and products in chemical reactions. If you know the atoms.

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Presentation transcript:

Stoichiometry In Solution Chemistry

Stochiometry involves calculating the amounts of reactants and products in chemical reactions. If you know the atoms or ions in a formula or a reaction, you can use stoichiometry to determine the amounts of atoms or ions that react.

The Concentration of Ions Problem Calculate the concentration (in mol/L) of chloride ions in the following solutions g of potassium chloride dissolved in 100 mL solution 26.5 g of calcium chloride dissolved in 150 mL solution A mixture of the two solutions (volumes are additive)

19.8 g of potassium chloride dissolved in 100 mL solution KCl  = g/mol 19.8 g KCl x 1 mol = mol KCL g KCl  K + + Cl - So for every KCl mole there is a Cl - mole

We calculated mol KCl so there is also mol Cl mol Cl - / L = 2.66 mol/L Cl -

26.5 g of calcium chloride dissolved in 150 mL solution CaCl 2  ( ) = g/mol 26.5 g CaCl 2 x 1 mol = mol CaCl g CaCl 2  Ca Cl - So for every CaCl 2 mole there is 2 Cl - moles

We calculated mol CaCl 2 so there is mol Cl mol Cl - / L = 3.19 mol/L Cl -

A mixture of the two solutions (volumes are additive) Total amount of Cl - moles is = Total volume is = L Total concentration of Cl - is mol / L = 2.98 mol/L

Mass Percent of Ions Problem Rhubard leaves contain relatively high concentrations of oxalate ions (C 2 O 4 2- ), which are poisonous and cause respiratory problems. To determine the percent of oxalate ions, a student measure a mass of leaves to be g. Then the students ground up the leaves and added calcium chloride solutions to precipitate out the calcium oxalate. The dried calcium oxalte was g. What was the mass percent of oxalate in the leaves?

Find molar mass of calcium oxalate Use the given mass of dried calcium oxalate and the molar mass to determine the number of moles Write a net ionic equation for the formation of calcium oxalate Use the coefficients to determine a ratio between calcium oxalate and oxalate Once you have moles of oxalate, find the molar mass and then determine the mass of oxalate. Using the determined mass of oxalate and the mass of the leaves, determine the mass percent of oxalate.

Find molar mass of calcium oxalate CaC 2 O 4  (12.011) + 4( ) = Use the given mass of dried calcium oxalate and the molar mass to determine the number of moles g x 1 mol= mol CaC 2 O g Write a net ionic equation for the formation of calcium oxalate Ca 2+ + C 2 O 4 2-  CaC 2 O 4

Use the coefficients to determine a ratio between calcium oxalate and oxalate One C 2 O 4 2- makes one CaC 2 O mol CaC 2 O 4 means mol C 2 O 4 2- Once you have moles of oxalate, find the molar mass and then determine the mass of oxalate. (2(12.011) + 4( ) = g/mol) mol C 2 O 4 2- x g = g 1 mol Using the determined mass of oxalate and the mass of the leaves, determine the mass percent of oxalate g/238.6 g x 100% = 0.160%

Finding Minimum Volume to Precipitate Aqueous solutions that contain silver ions are usually treated with chloride ions to recover silver chloride. What is the minimum volume of 0.25 mol/L magnesium chloride needed to precipitate all the silver ions in 60 mL of 0.30 mol/L silver nitrate? (Assume that silver chloride is completely insoluble in water).

Find moles of silver chloride from the volume and concentration. Write and balance a chemical equation for the reaction Use the mole ratio between silver chloride and magnesium chloride to determine how much magnesium chloride is needed. Use that amount and the concentration of the solution to determine the volume needed.

Find moles of silver chloride from the volume and concentration. 60 mL of 0.30 mol/L L x 0.30 mol = mol AgNO 3 1 L Write and balance a chemical equation for the reaction MgCl AgNO 3  Mg(NO 3 ) AgCl Use the mole ratio between silver chloride and magnesium chloride to determine how much magnesium chloride is needed 1 MgCl 2 to 2 AgNO 3 (so mol MgCl 2 )

Use that amount and the concentration of the solution to determine the volume needed mol MgCl mol/L mol = L of MgCl 2 needed mol/L