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stoichiometry

Solutions. stoichiometry. 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:

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stoichiometry

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  1. Solutions stoichiometry

  2. 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) • Read pg. 351-353. Try Q 1-3a.

  3. 2.20 mol NH3 1 mol H2SO4 x x L NH3 2 mol NH3 0.02684 mol H2SO4 = Pg. 353, Question 1 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

  4. 0.125 mol Al2(SO4)3 3 mol Ca(OH)2 L Ca(OH)2 x x x L Al2(SO4)3 0.0250 mol Ca(OH)2 1 mol Al2(SO4)3 Pg. 353, Question 2 Calcium hydroxide is sometimes used in water treatment plants to clarify water for residential use. 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 LAl2(SO4)3 = 0.375 L Ca(OH)2

  5. 0.200 mol FeCl3 L Na2CO3 3 mol Na2CO3 x x x L FeCl3 0.250 mol Na2CO3 2 mol FeCl3 Pg. 353, Question 3 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 LFeCl3 = 0.0900 L Na2CO3 = 90.0 mL Na2CO3

  6. 0.125 mol Al2(SO4)3 2.20 mol NH3 3 mol Ca(OH)2 L Ca(OH)2 1 mol H2SO4 x x x x x L Al2(SO4)3 L NH3 0.0250 mol Ca(OH)2 1 mol Al2(SO4)3 2 mol NH3 0.02684 mol H2SO4 = Answers # mol H2SO4= 1. H2SO4(aq) + 2NH3(aq)  (NH4)2SO4(aq) Calculate mol H2SO4, then mol/L = mol/0.0500 L 0.0244 L NH3 mol/L = 0.02684 mol H2SO4/ 0.0500 L = 0.537 mol/L 2. Al2(SO4)3(aq)+3Ca(OH)2(aq)2Al(OH)3(s)+3CaSO4(s) # L Ca(OH)2= 0.0250 LAl2(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 3. 2FeCl3(aq) + 3Na2CO3(aq)  Fe2(CO3)3(s) + 6NaCl(aq) Answers # L Na2CO3= 0.0750 LFeCl3 = 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 LNaOH 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 1 mol Ag3PO4 3 mol AgNO3 418.58 g Ag3PO4 L AgNO3 99.40 g Zn(OH)2 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 LNaOH 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 LK3PO4 4b. 3AgNO3(aq)+K3PO4(aq)Ag3PO4(s) +3KNO3(aq) # g Ag3PO4= = 5.2 g 0.025 LK3PO4

  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 NaOH L Al(NO3)3 1 mol Al(OH)3 1 mol Al(OH)3 # g Al(OH)3= 0.550 LAl(NO3)3 5. Al(NO3)3(aq) + 3NaOH(aq)  Al(OH)3(s) + 3NaNO3(aq) # g Al(OH)3= 0.240 L NaOH For more lessons, visit www.chalkbored.com

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