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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. 1 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. 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) + ? grams 3.45 g 3 H2(g) Units match 3.45 g Al m ol Al 27.0g Al 2 m ol AlC l3 133.3g Al C l3 mol Al C l3 2 m ol 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 3 4 Molarity Molarity is a term used to express concentration. The units of molarity are moles per liter (It is abbreviated as a capital M) When working problems, it is a good idea to change M into its units. moles moles M Liter 1000 mL 5 6 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. 7 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 m ol = 0.140 mol 3 133.4g 200.0 x 10 L L molar mass of AlCl3 2nd: M1V1 = M2V2 dilution formula (0.140 M)(10.0 mL) = (? M)(100.0 mL) 0.0140 M = M2 final concentration 8 9 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 0.125mol 0.125mol L 1000m L 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 0.125mol 0.125mol L 1000m L 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 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? 1st write out a balanced chemical equation 14 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 2H2O(l) + BaCl2 Units match HCl 2 mol 1000 mL Ba(OH)2 1 mol Ba(OH)2 HCl 1000 mL 0.40 mol HCl = 176 mL HCl 15 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 0.135 mol L ? mol L First write a balanced chemical reaction. 17 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 23.28mLHCl 25.00 x 10-3 L Ba(OH)2 0.135mol HCl l molBa(OH)2 = 0.0629 mol Ba(OH)2 1000mLHCl 2 molHCl L Ba(OH)2 Units Already Match on Bottom! 18 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. We must first write a balanced equation. 20 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 0.385m ol1molC a(O H)2 =0.0770 mol(Ca(OH) 1000m L 2molHNO3 48.0 x 10-3L L (Ca(OH) ) 2) 2 HNO 3 units match! 21 22 Limiting/Excess/ Reactant and Theoretical Yield Problems : Potassium superoxide, KO2, is used in rebreathing gas masks to generate oxygen. 4KO2(s) + 2H2O(l) 4KOH(s) + 3O2(g) a. How many moles of O2 can be produced from 0.15 mol KO2 and 0.10 mol H2O? b. Determine the limiting reactant. 4KO2(s) + 2H2O(l) 4KOH(s) + 3O2(g) 0.15 mol ? moles 0.10 mol Hide Two starting amounts? Where do we start? one 23 Try this problem (then check your answer): Calculate the molarity of a solution prepared by dissolving 25.6 grams of Al(NO3)3 in 455 mL of solution. After you have worked the problem, click here to see setup answer 25.6g mole mol 0.264 -3 213g 455 x 10 L L 24 25