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S OLUTION STOICHIOMETRY Problems and solutions

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V OCAB : 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 - H 2 SO 4 H + + HSO 4 -

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E XAMPLE #1 D ILUTION If you need to make 75ml of 1.5M HCl solution from 12M stock HCl. How much stock solution and water are needed? M 1 V 1 = M 2 V 2 75ml(1.5M) = (12M)(V 2 ) 9.4ml of stock and 65.6ml H 2 O

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E XAMPLE #2 S TOICHIOMETRY What volume of Na 3 PO 4 (6.0M) is needed to react with 120ml of 4.2M ZnCl 2 ? Both reactants are soluble: Na 3 PO 4 3Na + + PO 4 3- ZnCl 2 2Cl - + Zn 2+ Net ionic: 3Zn 2+ + 2PO 4 3- Zn 3 (PO 4 ) 2

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E XAMPLE #2… CONT. Net ionic: 3Zn 2+ + 2PO 4 3- Zn 3 (PO 4 ) 2 Use stoichiometry:.120L 4.2molZnCl 2 1molZn 2+ 2molPO 4 3- 1molNa 3 PO 4 1L = 1L 1molZnCl 2 3molZn 2+ 1molPO 4 3- 6molNa 3 PO 4 =.056L = 56ml Na 3 PO 4

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E XAMPLE #3 A 1.42g sample of a pure compound, with the formula M 2 SO 4, 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

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E XAMPLE #3 – J UST ONE METHOD OF SOLVING M 2 SO 4 + CaCl 2 2MCl + CaSO 4 1.42g 1.36g 1.36gCaSO 4 1molCaSO 4 1molM 2 SO 4 Xg = 136g 1molCaSO 4 1mol M 2 SO 4.01x g = 1.42g M 2 SO 4 X=142g M 2 SO 4 subtract the mass of SO 4 2M = 45.94g therefore, mass M = 22.97g which is Na

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E XAMPLE #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.

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E XAMPLE #4 – J UST ONE METHOD OF SOLVING MgCl 2 + 2AgNO 3 2AgCl + Mg(NO 3 ) 2 1.50g mixture 0.500M.641g.641gAgCl 1molAgCl 1molMgCl 2 95.21g =.213gMgCl 2 143.15g 2mol AgCl 1molMgCl 2.213gMgCl 2 / 1.50g mixture = 14.2%.213gMgCl 2 1molMgCl 2 2molAgNO 3 1L = 8.95ml AgNO 3 95.21g 1molMgCl 2.500M AgNO 3

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E XAMPLE #5 A 230. ml sample of 0.275M CaCl 2 solution is left on a hotplate overnight. The following morning, the solution is 1.10M. What volume of water evaporated from the 0.275M CaCl 2 solution?

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E XAMPLE #5 – J UST ONE METHOD OF SOLVING CaCl 2.275M 230ml (.230L).275M = Xmol /.230L.0633mol CaCl 2 1.10M =.0633mol / XL.0575L = vol in morning 230ml-57.5ml = 173ml evaporated

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E XAMPLE #6 M IXING Given 120ml of 2.2M LiCl and 400ml of 4.2M AlCl 3, what is the concentration of Cl - after mixing? 1. Write dissociation reactions: LiCl Li + + Cl - AlCl 3 Al 3+ + 3Cl - 2. Determine the moles of Cl - in each reaction:.120L 2.2mol LiCl 1mol Cl - = 0.264 mol Cl - 1L 1mol LiCl

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E XAMPLE #6…. CONT..400L 4.2mol AlCl 3 3mol Cl - = 5.04mol Cl - 1L 1mol AlCl 3 3. Add moles of Cl - together = 5.304 mol Cl - 4. Divide moles by the total volume of the solution. 5.304mol Cl - /.520L = 10.2M

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Solutions Solubility -the amount of solute that can be dissolved to form a solution. Solvent – the substance in a solution present in the greatest amount.

Solutions Solubility -the amount of solute that can be dissolved to form a solution. Solvent – the substance in a solution present in the greatest amount.

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