Solution Stoichiometry 1 Solution Stoichiometry In stoichiometric calculations, molarity is used to calculate moles from volume of solution analogous to using molar mass from mass of a solid. It can also be related to the volume at STP 2 Gram to Gram Conversions Aluminum is an active metal that when placed in hydrochloric acid produces hydrogen gas and aluminum chloride. How many grams of aluminum chloride can be produced when 3.45 grams of aluminum are reacted with an excess of hydrochloric acid? 2 Al(s) + 6HCl(aq) 2AlCl3(aq) + 3 H2(g) First write a balanced equation. 3 Gram to Gram Conversions Aluminum is an active metal that when placed in hydrochloric acid produces hydrogen gas and aluminum chloride. How many grams of aluminum chloride can be produced when 3.45 grams of aluminum are reacted with an excess of hydrochloric acid? 2 Al(s) + 6HCl(aq) 2AlCl3(aq) + ? grams 3.45 g 3 H2(g) Now let’s get organized. Write the information below the substances. 4 gram to gram conversions Aluminum is an active metal that when placed in hydrochloric acid produces hydrogen gas and aluminum chloride. How many grams of aluminum chloride can be produced when 3.45 grams of aluminum are reacted with an excess of hydrochloric acid? 2 Al(s) + 6HCl(aq) 2AlCl3(aq) + ? grams 3.45 g 3 H2(g) Units match 3.45 g Al mol Al 27.0 g Al 2 mol AlCl 3 133.3 g AlCl 3 mol AlCl 3 2 mol Al Now We must Now use Let’s the always use work molar thethe convert molar mass problem. ratio. to toconvert moles. to grams. = 17.0 g AlCl3 5 6 7 Solution Stoichiometry: Determine how many mL of 0.102 M NaOH solution are needed to neutralize 35.0 mL of 0.125 M H2SO4 solution. 2 1 2SO4 ____NaOH + ____H 2 2O ____H 1 2SO4 + ____Na First write a balanced Equation. 10 Solution Stoichiometry: Determine how many mL of 0.102 M NaOH solution is needed to neutralize 35.0 mL of 0.125 M H2SO4 solution. 2 1 2SO4 ____NaOH + ____H 0.102 M mol L ? mL Our Goal 2 2O ____H 1 2SO4 + ____Na 35.0 mL 0.125 mol 0.125 mol L 1000 mL Since 1 L = 1000 mL, we can use this to save on the number of conversions Now, let’s get organized. Place numerical Information and accompanying UNITS below each compound. 11 Solution Stoichiometry: Determine how many mL of 0.102 M NaOH solution is needed to neutralize 35.0 mL of 0.125 M H2SO4 solution. 2 1 2SO4 ____NaOH + ____H 0.102 M mol L ? mL H2SO4 35.0 mL 2 2O ____H 1 2SO4 + ____Na 35.0 mL 0.125 mol 0.125 mol L 1000mL H2SO4 0.125 mol 1000 mL H2SO4 NaOH 2 mol 1 mol H2SO4 Units Match 1000 mL NaOH = 85.8 mL NaOH 0.102 mol NaOH Now let’s get to work converting. 12 Solution Stoichiometry What volume of 0.40 M HCl solution is needed to completely neutralize 47.1 mL of 0.75 M Ba(OH)2? 1st write out a balanced chemical equation 13 Solution Stoichiometry What volume of 0.40 M HCl solution is needed to completely neutralize 47.1 mL of 0.75 M Ba(OH)2? 2HCl(aq) + Ba(OH)2(aq) 0.40 M 47.1 mL 0.75 M ? mL Ba(OH)2 47.1 mL 0.75mol Ba(OH)2 1000 mL Ba(OH)2 2H2O(l) + BaCl2 Units match HCl 2 mol 1 mol Ba(OH)2 HCl 1000 mL 0.40 mol HCl = 176 mL HCl 14 15 Solution Stochiometry Problem: A chemist performed a titration to standardize a barium hydroxide solution. If it took 23.28 mL of 0.135 M hydrochloric acid to neutralize 25.00 mL of the barium hydroxide solution, what was the concentration of the barium hydroxide solution in moles per liter (M)? 2 1 2 2O(l) + ____BaCl 1 ____HCl(aq) + ____Ba(OH) 2(aq) ____H 2(aq) 25.00 mL 23.28 mL 0.135 mol L ? mol L First write a balanced chemical reaction. 16 Solution Stochiometry Problem: A chemist performed a titration to standardize a barium hydroxide solution. If it took 23.28 mL of 0.135 M hydrochloric acid to neutralize 25.00 mL of the barium hydroxide solution, what was the concentration of the barium hydroxide solution in moles per liter (M)? 2 1 2 2O(l) + ____BaCl 1 ____HCl(aq) + ____Ba(OH) 2(aq) ____H 2(aq) 25.00 mL 23.28 mL Units match on top! ? mol 0.135 mol L L 23.28 mL HCl 25.00 x 10-3 L Ba(OH)2 0.135 mol HCl l mol Ba(OH)2 = 0.0629 mol Ba(OH)2 1000 mL HCl 2 mol HCl L Ba(OH) 2 Units Already Match on Bottom! 17 18 Solution Stochiometry Problem: 48.0 mL of Ca(OH)2 solution was titrated with 19.2 mL of 0.385 M HNO3. Determine the molarity of the Ca(OH)2 solution. We must first write a balanced equation. 19 Solution Stochiometry Problem: 48.0 mL of Ca(OH)2 solution was titrated with 19.2 mL of 0.385 M HNO3. Determine the molarity of the Ca(OH)2 solution. Ca(OH)2(aq) + 2 HNO3(aq) 2 H2O(l) + Ca(NO3)2(aq) 48.0 mL ?M HNO3 19.2 mL 19.2 mL 0.385 M 0.385 mol L HNO 3 0.385 mol 1mol Ca(OH) 2 =0.0770 mol(Ca(OH) 1000 mL 2mol HNO 3 48.0 x 10-3L L (Ca(OH) ) 2) 2 HNO 3 units match! 20 21 Titration - is the process where the concentration of an unknown solution is determined by reacting It with a solution with known concentration. Dilution – dilution add water to the solution of known Volume and concentration. The number of mole of solute Does not change in dilution because the amount of solute Do not change. mole solute concentrated solution = mole solute dilute solution V1C1 = V2C2 22 Solutions A solution is prepared by dissolving 3.73 grams of AlCl3 in water to form 200.0 mL solution. A 10.0 mL portion of the solution is then used to prepare 100.0 mL of solution. Determine the molarity of the final solution. What type of problem(s) is this? Molarity followed by dilution. 23 Solutions A solution is prepared by dissolving 3.73 grams of AlCl3 in water to form 200.0 mL solution. A 10.0 mL portion of the solution is then used to prepare 100.0 mL of solution. Determine the molarity of the final solution. 1st: 3.73 g mol = 0.140 mol 3 133.4 g 200.0 x 10 L L molar mass of AlCl3 2nd: C1V1 = C2V2 dilution formula (0.140 M)(10.0 mL) = (? C2)(100.0 mL) 0.0140 M = C2 final concentration 24 4 mL of a sample was diluted to 100 mL in a volumetric flask. The diluted solution was analyzed and found to have a concentration of 2.0 mg/L. What was the concentration of the original sample? C1 x V1 = C2 x V2 C1 = ? V1 = 4 mL C2 = 2.0 mg/L V2 = 100 mL C1 x 4 mL = 2.0 mg/L x 100 mL 4 mL 4 mL C1= 2.0 x 100 mg/L 4 50.0 mg/L = C1