1 / 12

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:

teneil
Télécharger la présentation

stoichiometry

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  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

More Related