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Solution stoichiometry

Solution stoichiometry. Problems and solutions. Vocab: the basics first. Strong Acids and Bases: Dissociate 100% in water Acids : CBS PIN HCl HBr H 2 SO 4 HClO 4 HI HNO 3 Bases : Group IA (Li-Cs) and IIA (Mg-Ba) hydroxides Mg(OH) 2 Mg 2+ + 2OH -

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Solution stoichiometry

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  1. Solution stoichiometry Problems and solutions

  2. Vocab: the basics first Strong Acids and Bases: Dissociate 100% in water Acids: CBS PIN HClHBr H2SO4 HClO4 HI HNO3 Bases: Group IA (Li-Cs) and IIA (Mg-Ba) hydroxides Mg(OH)2 Mg2+ + 2OH- H2SO4 H+ + HSO4-

  3. Example #1 Dilution If you need to make 75ml of 1.5M HCl solution from 12M stock HCl. How much stock solution and water are needed? M1V1 = M2V2 75ml(1.5M) = (12M)(V2) 9.4ml of stock and 65.6ml H2O

  4. Example #2 Stoichiometry What volume of Na3PO4 (6.0M) is needed to react with 120ml of 4.2M ZnCl2? Both reactants are soluble: Na3PO4 3Na+ + PO43- ZnCl2 2Cl- + Zn2+ Net ionic: 3Zn2+ + 2PO43- Zn3(PO4)2

  5. Example #2…cont. Net ionic: 3Zn2+ + 2PO43- Zn3(PO4)2 Use stoichiometry: .120L 4.2molZnCl2 1molZn2+ 2molPO43- 1molNa3PO4 1L = 1L 1molZnCl2 3molZn2+ 1molPO43- 6molNa3PO4 =.056L = 56ml Na3PO4

  6. Example #3 A 1.42g sample of a pure compound, with the formula M2SO4, was dissolved in water and treated with an excess of aqueous calcium chloride, resulting in the precipitation of all the sulfate ions as calcium sulfate. The precipitate was collected and found to have a mass of 1.36g. Determine the atomic mass of M and identify M

  7. Example #3 – Just one method of solving M2SO4 + CaCl2 2MCl + CaSO4 1.42g 1.36g 1.36gCaSO4 1molCaSO4 1molM2SO4Xg = 136g 1molCaSO41mol M2SO4 .01x g = 1.42g M2SO4 X=142g M2SO4 subtract the mass of SO4 2M = 45.94g therefore, mass M = 22.97g which is Na

  8. Example #4 Consider a 1.50g mixture of magnesium nitrate and magnesium chloride. After dissolving this mixture in water, 0.500M silver nitrate is added drop-wise until precipitate formation is complete. The mass of the white precipitate formed is 0.641g. • Calculate the mass percent of magnesium chloride in the mixture • Determine the minimum volume of silver nitrate that must be added to ensure the complete formation of the precipitate.

  9. Example #4 – Just one method of solving MgCl2 + 2AgNO3 2AgCl + Mg(NO3)2 1.50g mixture 0.500M .641g .641gAgCl 1molAgCl 1molMgCl2 95.21g = .213gMgCl2 143.15g 2mol AgCl 1molMgCl2 .213gMgCl2 / 1.50g mixture = 14.2% .213gMgCl2 1molMgCl2 2molAgNO3 1L = 8.95ml AgNO3 95.21g 1molMgCl2 .500M AgNO3

  10. Example #5 A 230. ml sample of 0.275M CaCl2 solution is left on a hotplate overnight. The following morning, the solution is 1.10M. • What volume of water evaporated from the 0.275M CaCl2 solution?

  11. Example #5 – Just one method of solving CaCl2 .275M 230ml (.230L) .275M = Xmol / .230L .0633mol CaCl2 1.10M = .0633mol / XL .0575L = vol in morning 230ml-57.5ml = 173ml evaporated

  12. Example #6 Mixing Given 120ml of 2.2M LiCl and 400ml of 4.2M AlCl3, what is the concentration of Cl- after mixing? • Write dissociation reactions: LiCl Li+ + Cl- AlCl3 Al3+ + 3Cl- • Determine the moles of Cl- in each reaction: .120L 2.2mol LiCl 1mol Cl- = 0.264 molCl- 1L 1mol LiCl

  13. Example #6….cont. .400L 4.2mol AlCl3 3mol Cl- = 5.04mol Cl- 1L 1mol AlCl3 • Add moles of Cl- together = 5.304 molCl- • Divide moles by the total volume of the solution. 5.304mol Cl- / .520L = 10.2M

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