Gram to Gram Conversions in Aluminum-Hydrochloric Acid Reaction

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? Al(s) + HCl(aq)  AlCl3(aq) + H2(g) 2 6 2 3 First write a balanced equation.

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? Al(s) + HCl(aq)  AlCl3(aq) + H2(g) 2 6 2 3 ? grams 3.45 g Now let’s get organized. Write the information below the substances.

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? Al(s) + HCl(aq)  AlCl3(aq) + H2(g) 2 6 2 3 ? grams 3.45 g Units match 3.45 g Al = g AlCl3 17.0 Let’s work the problem. We must always convert to moles. Now use the molar ratio. Now use the molar mass to convert to grams.

5. 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.

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. What type of problem(s) is this? Molarity followed by dilution.

8. 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 L 0.140 200.0 x 10-3 L molar mass of AlCl3 dilution formula M1V1 = M2V2 2nd: (0.140 M)(10.0 mL) = (? M)(100.0 mL) final concentration 0.0140 M = M2

10. 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 2 1 ____NaOH + ____H2SO4 ____H2O + ____Na2SO4 First write a balanced Equation.

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 2 1 ____NaOH + ____H2SO4 ____H2O + ____Na2SO4 0.102 M 35.0 mL ? mL Our Goal 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.

12. 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 2 1 ____NaOH + ____H2SO4 ____H2O + ____Na2SO4 0.102 M 35.0 mL ? mL Units Match shortcut H2SO4 35.0 mL H2SO4 0.125 mol 1000 mL H2SO4 NaOH 2 mol 1 mol H2SO4 1000 mL NaOH 0.102 mol NaOH = mL NaOH 85.8 Now let’s get to work converting.

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? 1st write out a balanced chemical equation

15. 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)  2H2O(l) + BaCl2 47.1 mL 0.75 M 0.40 M ? mL Units match HCl 2 mol Ba(OH)2 47.1 mL HCl 1000 mL 176 = mL HCl 0.40 mol HCl 1 mol Ba(OH)2

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 1 ____HCl(aq) + ____Ba(OH)2(aq)  ____H2O(l) + ____BaCl2(aq) 25.00 mL 23.28 mL ? mol L 0.135 mol L First write a balanced chemical reaction.

18. 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 1 ____HCl(aq) + ____Ba(OH)2(aq)  ____H2O(l) + ____BaCl2(aq) 25.00 mL 23.28 mL Units match on top! ? mol L 0.135 mol L = mol Ba(OH)2 L Ba(OH)2 0.0629 25.00 x 10-3 L Ba(OH)2 Units Already Match on Bottom!

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. We must first write a balanced equation.

21. 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) + HNO3(aq)  2 H2O(l) 2 + Ca(NO3)2(aq) 48.0 mL 19.2 mL ? M 0.385 M HNO3 19.2 mL mol(Ca(OH)2) L (Ca(OH)2) = 0.0770 48.0 x 10-3L units match!

23. 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 one Two starting amounts? Where do we start?

24. 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