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stoichiometry

Solutions. stoichiometry.

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stoichiometry

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  1. Solutions stoichiometry

  2. Solution types of stoichiometry problems are no harder than any other stoichiometry problem. You must use the concentration given to convert to moles, then use the reaction 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

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

  4. 1 mol Cu 2 mol AgNO3 x x 63.5 g Cu 1 mol Cu 0.38 mol AgNO3 = 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 # mol AgNO3 = 12.0 g Cu L = 0.38 mol AgNO3/ 2.00 mol / L = 0.19L = 190mL

  5. 2.20 mol NH3 1 mol H2SO4 x x L NH3 2 mol NH3 = 0.02684 mol H2SO4 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 # mol H2SO4= 0.0244 L NH3 mol/L = 0.02684 mol H2SO4/ 0.0500 L = 0.537 mol/L

  6. 0.125 mol Al2(SO4)3 L Ca(OH)2 3 mol Ca(OH)2 x x x L Al2(SO4)3 0.0250 mol Ca(OH)2 1 mol Al2(SO4)3 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 = 0.375 L Ca(OH)2

  7. 0.200 mol FeCl3 L Na2CO3 3 mol Na2CO3 x x x L FeCl3 0.250 mol Na2CO3 2 mol FeCl3 A chemistry teacher wants 75.0 mL of 0.200 mol/L iron(Ill) chloride solution to react completely with an excess of 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= 0.0750 L FeCl3 = 0.0900 L Na2CO3 = 90.0 mL Na2CO3

  8. Assignment • 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.

  9. 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 precipitate is produced from the above reaction? • What mass of precipitate should result when 0.550 L of 0.500 mol/L aluminum nitrate solution is mixed with 0.240 L of 1.50 mol/L sodium hydroxide solution?

  10. 0.50 mol NaOH 0.600 mol Fe2(SO4)3 L H2SO4 1 mol H2SO4 2 mol Fe(OH)3 x x x x x L NaOH L Fe2(SO4)3 2.0 mol H2SO4 1 mol Fe2(SO4)3 2 mol NaOH Answers # L H2SO4= 0.075 L NaOH 1. H2SO4(aq) + 2NaOH(aq)  2H2O + Na2SO4(aq) = 0.009375 L = 9.4 mL 2. Fe2(SO4)3(aq) + 6NaOH(aq)  2Fe(OH)3(s) + 3Na2SO4(aq) # mol Fe(OH)3= 85 L Fe2(SO4)3 = 102 mol

  11. 0.50 mol K3PO4 0.50 mol K3PO4 2.50 mol NaOH 1 mol Zn(OH)2 418.58 g Ag3PO4 L AgNO3 99.40 g Zn(OH)2 1 mol Ag3PO4 3 mol AgNO3 x x x x x x x x x L K3PO4 L K3PO4 L NaOH 1 mol Zn(OH)2 1 mol Ag3PO4 0.20 mol AgNO3 1 mol K3PO4 1 mol K3PO4 2 mol NaOH # g Zn(OH)2= = 6.21 g 0.0500 L NaOH 3. 2NaOH(aq) + ZnCl2(aq)  Zn(OH)2(s) + 2NaCl(aq) 4a. 3AgNO3(aq) + K3PO4(aq)  Ag3PO4(s) + 3KNO3(aq) # L AgNO3 = = 0.1875 L = 0.19 L 0.025 L K3PO4 4b. 3AgNO3(aq) + K3PO4(aq)  Ag3PO4(s) + 3KNO3(aq) # g Ag3PO4= = 5.2 g 0.025 L K3PO4

  12. 1 mol Al(OH)3 1 mol Al(OH)3 x x 1 mol Al(NO3)3 3 mol NaOH = = 21.4 g Al(OH)3 9.36 g Al(OH)3 1.50 mol NaOH 0.500 mol Al(NO3)3 77.98 g Al(OH)3 77.98 g Al(OH)3 x x x x L Al(NO3)3 L NaOH 1 mol Al(OH)3 1 mol Al(OH)3 # g Al(OH)3= 0.550 L Al(NO3)3 5. Al(NO3)3(aq) + 3NaOH(aq)  Al(OH)3(s) + 3NaNO3(aq) # g Al(OH)3= 0.240 L NaOH

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