1. Stoichiometry with Solutions Combining what you’ve just learned with what you are already good at

2. Stoichiometry with Solutions Now that you know about concentration, you have another fact you can use to work with moles. Because Molarity = moles/Liter, this can be used as a conversion factor. For example, how many moles of NaCl are in 0.25L of a 2.0M solution? Remember, the Molarity gives you your conversion factor. Set up the problem using dimensional analysis: 0.25 L x 2.0 mol = 0.50 mol NaCl 1.0 L Write molarity as a fraction

3. Try another How many moles of iron (III) chloride are in 0.32 L of a 1.5 M solution? 0.32 L x 1.5 mol = 0.48 mol FeCl 3 1.0 L And one more – how many moles of silver nitrate are in 0.50 L of a 0.05 M solution? 0.50 L x 0.05 mol = 0.025 mol AgNO 3 1.0 L

4. Once you have determined the number of moles, you can do any other conversion you already know about moles (mass, mole ratio in a reaction, etc). That means – we have an even-more-improved mole map!

5. Even-more-improved mole map Molarity written as a fraction

6. Remember – to get from volume of a solution to moles, you need the concentration (the Molarity)! Consider the following reaction: AgNO 3 (aq) + HCl (aq)  AgCl (s) + HNO 3 (aq) If 0.1 L of a 0.1 M solution of AgNO 3 is mixed with unlimited HCl, how many grams of AgCl will be made? 0.1 L x 0.1 mol AgNO 3 x 1 mol AgCl x 143.32 g AgCl = 1.4332 g AgCl 1.0 L 1 mol AgNO 3 1 mol AgCl Trace the path taken on the mole map above.

7. Try another one: If 0.2 L of a 1.0 M solution of HCl react with unlimited silver nitrate, how many grams of AgCl will be made? 0.2 L x 1.0 M HCl x 1 mol AgCl x 143.32 g AgCl = 28.664 g AgCl 1.0 L 1 mol HCl 1 mol AgCl

8. And one more: A solution of 0.2 M iron (III) chloride reacts with aqueous sodium hydroxide to produce a precipitate of iron (III) hydroxide and aqueous sodium chloride. Write and balance the equation. FeCl 3 (aq) + 3 NaOH (aq)  Fe(OH) 3 (s) + 3 NaCl (aq)

9. If 0.025 L of the iron (III) chloride solution reacts with an unlimited NaOH, how many grams of precipitate will be made? 0.025 L FeCl 3 x 0.2 mol FeCl 3 x 1 mol Fe(OH) 3 x 106.874 g Fe(OH) 3 = 0.53437 g Fe(OH) 3 1.0 L 1 mol FeCl 3 1 mol Fe(OH) 3