Stoichiometry with Solutions

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

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

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 FeCl3 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 AgNO3

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!

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

AgNO3 (aq) + HCl (aq)  AgCl (s) + HNO3 (aq) Remember – to get from volume of a solution to moles, you need the concentration (the Molarity)! Consider the following reaction: AgNO3 (aq) + HCl (aq)  AgCl (s) + HNO3 (aq) If 0.1 L of a 0.1 M solution of AgNO3 is mixed with unlimited HCl, how many grams of AgCl will be made? 0.1 L x 0.1 mol AgNO3 x 1 mol AgCl x 143.32 g AgCl = 1.4332 g AgCl 1.0 L 1 mol AgNO3 1 mol AgCl Trace the path taken on the mole map above.

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 mol HCl x 1 mol AgCl x 143.32 g AgCl = 28.664 g AgCl 1.0 L 1 mol HCl 1 mol AgCl

FeCl3 (aq) + 3 NaOH (aq)  Fe(OH)3 (s) + 3 NaCl (aq) 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. FeCl3 (aq) + 3 NaOH (aq)  Fe(OH)3 (s) + 3 NaCl (aq)

FeCl3 (aq) + 3 NaOH (aq)  Fe(OH)3 (s) + 3 NaCl (aq) 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 FeCl3 x 0.2 mol FeCl3 x 1 mol Fe(OH)3 x 106.874 g Fe(OH)3 = 0.53437 g Fe(OH)3 1.0 L 1 mol FeCl3 1 mol Fe(OH)3