Solutions Stoichiometry

Solution types of stoichiometry problems are no harder than any other stoichiometry problem. You must use the concentration given (molarity) to convert to moles, then use the molar ratio to convert to moles of the final product, then use concentration of the final product to convert to the final answer. Moles/L Moles/L

Stoichiometry overview Recall that in stoichiometry the mole ratio provides a necessary conversion factor: grams (x)  moles (x)  moles (y)  grams (y) molar mass of x molar mass of y mole ratio from balanced equation We can do something similar with solutions: volume (x)  moles (x)  moles (y)  volume (y) mol/L of x mol/L of y mole ratio from balanced equation

Cu(s) + 2AgNO3(aq)  Cu(NO3)2 + 2Ag(s) Calculate the volume of a 2.00 mol/L silver nitrate solution that is needed for 12.0 g of copper metal to react according to the following equation? Cu(s) + 2AgNO3(aq)  Cu(NO3)2 + 2Ag(s) Calculate the moles of AgNO3, then volume of AgNO3 2 mol AgNO3 1 mol Cu x 1 L 1 mol Cu 63.5 g Cu x X 12.0 g Cu 2 mol = 0.19L = 190mL

Al2(SO4)3(aq) + 3Ca(OH)2(aq)  2Al(OH)3(s) + 3CaSO4(s) Calculate the volume of 0.0250 mol/L calcium hydroxide solution that can be completely reacted with 25.0 mL of 0.125 mol/L aluminum sulfate solution. Al2(SO4)3(aq) + 3Ca(OH)2(aq)  2Al(OH)3(s) + 3CaSO4(s) # L Ca(OH)2= 0.0250 L Al2(SO4)3 0.125 mol Al2(SO4)3 L Al2(SO4)3 x 3 mol Ca(OH)2 1 mol Al2(SO4)3 x L Ca(OH)2 0.0250 mol Ca(OH)2 x = 0.375 L Ca(OH)2

H2SO4(aq) + 2NH3(aq)  (NH4)2SO4(aq) Ammonium sulfate is manufactured by reacting sulfuric acid with ammonia. What concentration of sulfuric acid is needed to react with 24.4 mL of a 2.20 mol/L ammonia solution if 50.0 mL of sulfuric acid is used? H2SO4(aq) + 2NH3(aq)  (NH4)2SO4(aq) Calculate mol H2SO4, then mol/L = mol/0.0500 L 2.20 mol NH3 L NH3 x 1 mol H2SO4 2 mol NH3 x 0.0244 L NH3 0.02684 mol H2SO4 = mol/L = 0.02684 / 0.0500 L = 0.537 mol/L

2FeCl3(aq) + 3Na2CO3(aq)  Fe2(CO3)3(s) + 6NaCl(aq) A chemistry teacher wants 75.0 mL of 0.200 mol/L iron(Ill) chloride solution to react completely with a 0.250 mol/L sodium carbonate solution. What volume of sodium carbonate solution is needed? 2FeCl3(aq) + 3Na2CO3(aq)  Fe2(CO3)3(s) + 6NaCl(aq) L Na2CO3 0.250 mol Na2CO3 x 0.0750 L FeCl3 0.200 mol FeCl3 L FeCl3 x 3 mol Na2CO3 2 mol FeCl3 x = 0.0900 L Na2CO3 = 90.0 mL Na2CO3

Assignment-Worksheet H2SO4 reacts with NaOH, producing water and sodium sulfate. What volume of 2.0 M H2SO4 will be required to react completely with 75 mL of 0.50 mol/L NaOH? How many moles of Fe(OH)3 are produced when 85.0 L of iron(III) sulfate at a concentration of 0.600 mol/L reacts with excess NaOH? What mass of precipitate will be produced from the reaction of 50.0 mL of 2.50 mol/L sodium hydroxide with an excess of zinc chloride solution.

Assignment a) What volume of 0.20 mol/L AgNO3 will be needed to react completely with 25.0 mL of 0.50 mol/L potassium phosphate? b) What mass of silver phosphate is produced from the above reaction?

Assignment 5. What mass of Al(OH)3 should result when 0.55 L of 0.50 mol/L aluminum nitrate [Al(NO3)3] solution is mixed with 0.24 L of 1.50 mol/L sodium hydroxide (NaOH) solution?

2. Fe2(SO4)3(aq) + 6NaOH(aq)  2Fe(OH)3(s) + 3Na2SO4(aq) Answers 1. H2SO4(aq) + 2NaOH(aq)  2H2O + Na2SO4(aq) # L H2SO4= 0.075 L NaOH 0.50 mol NaOH L NaOH x 1 mol H2SO4 2 mol NaOH x L H2SO4 2.0 mol H2SO4 x = 0.009375 L = 9.4 mL 2. Fe2(SO4)3(aq) + 6NaOH(aq)  2Fe(OH)3(s) + 3Na2SO4(aq) # mol Fe(OH)3= 0.600 mol Fe2(SO4)3 L Fe2(SO4)3 x 2 mol Fe(OH)3 1 mol Fe2(SO4)3 x 85 L Fe2(SO4)3 = 102 mol

3. 2NaOH(aq) + ZnCl2(aq)  Zn(OH)2(s) + 2NaCl(aq) # g Zn(OH)2= 0.0500 L NaOH 2.50 mol NaOH L NaOH x 1 mol Zn(OH)2 2 mol NaOH x 99.40 g Zn(OH)2 1 mol Zn(OH)2 x 4a. 3AgNO3(aq) + K3PO4(aq)  Ag3PO4(s) + 3KNO3(aq) # L AgNO3 = = 0.1875 L = 0.19 L 0.025 L K3PO4 0.50 mol K3PO4 L K3PO4 x 3 mol AgNO3 1 mol K3PO4 x L AgNO3 0.20 mol AgNO3 x 4b. 3AgNO3(aq) + K3PO4(aq)  Ag3PO4(s) + 3KNO3(aq) # g Ag3PO4= = 5.2 g 0.025 L K3PO4 0.50 mol K3PO4 L K3PO4 x 1 mol Ag3PO4 1 mol K3PO4 x 418.58 g Ag3PO4 1 mol Ag3PO4 x

0.5 mol Al(NO3)3 1L 1 mol Al(OH)3 78 g Al(OH)3 1 mol Al(OH)3 5. Al(NO3)3 + 3NaOH  3Na(NO3) + Al(OH)3 0.55 L Al(NO3)3 0.5 mol Al(NO3)3 1L 1 mol Al(OH)3 78 g Al(OH)3 Al(NO3)3 1 mol Al(NO3)3 1 mol Al(OH)3 Al(OH)3 = 21.45 grams