2. Stoichiometry

3. Stoichiometry Consider the chemical equation: 4NH 3 + 5O 2  6H 2 O + 4NO There are several numbers involved. What do they all mean? “stochio” = element “metry” = measurement Stoichiometry is about measuring the amounts of elements and compounds involved in a reaction.

4. Stoichiometry 4NH 3 + 5O 2  6H 2 O + 4NO 4 : 5 : 6 : 4 is the mole ratio.

5. Stoichiometry 4NH 3 + 5O 2  6H 2 O + 4NO 4 moles of NH 3 react with 5 moles of O 2 to produce 6 moles of H 2 O and 4 moles of NO

6. 5 Stoichiometry (more working with ratios) Ratios are found within a chemical equation. 2HCl + Ba(OH) 2  2H 2 O + BaCl 2 1 1 2 moles of HCl react with 1 mole of Ba(OH) 2 to form 2 moles of H 2 O and 1 mole of BaCl 2 coefficients give MOLAR RATIOS

7. STOICHIOMETRY MASS A VOLUME A MOL “B” MASS B VOLUME B MOL “A”

8. Moving along the stoichiometry path We always use the same type of information to make the jumps between steps: grams (x)  moles (x)  moles (y)  grams (y) Molar mass of x Molar mass of y Mole ratio from balanced equation © D Scott; CHS

9. Many stoichiometry problems follow a pattern: grams(x)  moles(x)  moles(y)  grams(y) Converting grams to grams We can start anywhere along this path depending on the question we want to answer Notice that we cannot directly convert from grams of one compound to grams of another. Instead we have to go through moles. © D Scott; CHS

10. Calculating Masses of Reactants and Products 1.Balance the equation. 2.Convert mass or volume to moles, if necessary. 3.Set up mole ratios. 4.Use mole ratios to calculate moles of desired material. 5.Convert moles to mass or volume, if necessary. 1.Balance the equation. 2.Convert mass or volume to moles, if necessary. 3.Set up mole ratios. 4.Use mole ratios to calculate moles of desired material. 5.Convert moles to mass or volume, if necessary.

11. Mole Ratios A mole ratio converts moles of one compound in a balanced chemical equation into moles of another compound.

12. 11 When N 2 O 5 is heated, it decomposes: 2N 2 O 5 (g)  4NO 2 (g) + O 2 (g) a. How many moles of NO 2 can be produced from 4.3 moles of N 2 O 5 ? = moles NO 2 4.3 mol N2O5N2O5 8.6 b. How many moles of O 2 can be produced from 4.3 moles of N 2 O 5 ? = mole O 2 4.3 mol N2O5N2O5 2.2 2N 2 O 5 (g)  4NO 2 (g) + O 2 (g) 4.3 mol? mol 2N 2 O 5 (g)  4NO 2 (g) + O 2 (g) 4.3 mol ? mol Mole – Mole Conversions Units match

13. 12 When N 2 O 5 is heated, it decomposes: 2N 2 O 5 (g)  4NO 2 (g) + O 2 (g) a. How many moles of N 2 O 5 were used if 210g of NO 2 were produced? = moles N 2 O 5 210 g NO 2 2.28 b. How many grams of N 2 O 5 are needed to produce 75.0 grams of O 2 ? = grams N 2 O 5 75.0 g O2O2 506 gram ↔ mole and gram ↔ gram conversions 2N 2 O 5 (g)  4NO 2 (g) + O 2 (g) 210g? moles 2N 2 O 5 (g)  4NO 2 (g) + O 2 (g) 75.0 g ? grams Units match

14. Example Reaction between magnesium and oxygen to form magnesium oxide. ( fireworks) 2 Mg(s) + O 2 (g) 2 MgO(s) Mole Ratios: 2 : 1: 2 How many moles of MgO is produced if 4 mols of Mg reacts with unlimited supply of O 2

15. 1) N 2 + 3 H 2 ---> 2 NH 3 a) How many moles of NH 3 if 6 moles of H 2 reacts with unlimited supply of N 2 ? b) How many moles of NH 3 is produced if 5 moles of H 2 reacts with unlimited supply of N 2 ? Practise Problems

16. 4NH 3 + 5O 2  6H 2 O + 4NO How many moles of H 2 O are produced if 2.00 moles of O 2 are used? Stoichiometry Question (1) 2.00 mol O 2 2.40 mol H 2 O = Notice that a correctly balanced equation is essential to get the right answer 6 mol H 2 O 5 mol O 2 © D Scott; CHS

17. 4 mol NO 6 mol H 2 O 4 NH 3 + 5 O 2  6 H 2 O + 4 NO How many moles of NO are produced in the reaction if 15 mol of H 2 O are also produced? Stoichiometry Question (2) 15 mol H 2 O 10. mol NO = © D Scott; CHS

18. 18.02 g H 2 O 1 mol H 2 O 6 mol H 2 O 4 mol NH 3 4 NH 3 + 5 O 2  6 H 2 O + 4 NO How many grams of H 2 O are produced if 2.2 mol of NH 3 are combined with excess oxygen? Stoichiometry Question (3) 2.2 mol NH 3 59 g H 2 O = © D Scott; CHS

19. 5 mol O 2 6 mol H 2 O 32 g O 2 1 mol O 2 4 NH 3 + 5 O 2  6 H 2 O + 4 NO How many grams of O 2 are required to produce 0.3 mol of H 2 O? Stoichiometry Question (4) 0.3 mol H 2 O 8 g O 2 = © D Scott; CHS

20. 4 NH 3 + 5 O 2  6 H 2 O + 4 NO How many grams of NO is produced if 12 g of O 2 is combined with excess ammonia? 4 mol NO 5 mol O 2 x Stoichiometry Question (5) 12 g O 2 9.0 g NO = 30.01 g NO 1 mol NO x 1 mol O 2 32 g O 2 x © D Scott; CHS

21. Have we learned it yet? Try these on your own - 4 NH 3 + 5 O 2  6 H 2 O + 4 NO a) How many moles of H 2 O can be made using 1.6 mol NH 3 ? b) what mass of NH 3 is needed to make 0.75 mol NO? c) how many grams of NO can be made from 47 g of NH 3 ? © D Scott; CHS

22. 4 NH 3 + 5 O 2  6 H 2 O + 4 NO a) b) c) Answers 6 mol H 2 O 4 mol NH 3 x 1.6 mol NH 3 2.4 mol H 2 O = 4 mol NH 3 4 mol NO x 0.75 mol NO 13 g NH 3 = 17.04 g NH 3 1 mol NH 3 x 4 mol NO 4 mol NH 3 x 47 g NH 3 83 g NO = 30.01 g NO 1 mol NO x 1 mol NH 3 17.04 g NH 3 x © D Scott; CHS

23. Mass-Mass Stoichiometry

24. 23 How many grams of aluminum chloride can be produced when 3.45 grams of aluminum are reacted with an excess of hydrochloric acid? First write a balanced equation. Al(s) + HCl(aq)  AlCl 3 (aq) + H 2 (g) 2 6 2 3 Gram to Gram Conversions

25. 24 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)  AlCl 3 (aq) + H 2 (g) 2 6 2 3 Now let’s get organized. Write the information below the substances. 3.45 g ? grams Gram to Gram Conversions

26. 25 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)  AlCl 3 (aq) + H 2 (g) 2 6 2 3 3.45 g ? grams Let’s work the problem. = g AlCl 3 3.45 g Al We must always convert to moles.Now use the molar ratio.Now use the molar mass to convert to grams. 17.0 Units match gram to gram conversions

27. Mass-Volume/Concentration Stoichiometry Just follow mass-mass problem to the penultimate level Convert moles of the substance into Volume.

28. Volume-Volume Stoichiometry Just follow mass-mass problem to the penultimate level Convert moles of the substance into Volume.

29. 28 50.0 mL of 6.0 M H 2 SO 4 (battery acid) were spilled and solid NaHCO 3 (baking soda) is to be used to neutralize the acid. How many grams of NaHCO 3 must be used? H 2 SO 4 (aq) + 2NaHCO 3  2H 2 O(l) + Na 2 SO 4 (aq) + 2CO 2 (g) Solution Stoichiometry

30. 29 50.0 mL 6.0 M ? g 50.0 mL of 6.0 M H 2 SO 4 (battery acid) were spilled and solid NaHCO 3 (baking soda) is to be used to neutralize the acid. How many grams of NaHCO 3 must be used? H 2 SO 4 (aq) + 2NaHCO 3  2H 2 O(l) + Na 2 SO 4 (aq) + 2CO 2 (g) Solution Stoichiometry = Our Goal = g NaHCO 3 H 2 SO 4 50.0 mL 1 mol H 2 SO 4 NaHCO 3 2 mol NaHCO 3 84.0 g mol NaHCO 3 50.4

31. 30 Solution Stoichiometry: Determine how many mL of 0.102 M NaOH solution are needed to neutralize 35.0 mL of 0.125 M H 2 SO 4 solution. First write a balanced Equation. ____NaOH + ____H 2 SO 4  ____H 2 O + ____Na 2 SO 4 2 1 2 1

32. 31 Solution Stoichiometry: Determine how many mL of 0.102 M NaOH solution is needed to neutralize 35.0 mL of 0.125 M H 2 SO 4 solution. Now, let’s get organized. Place numerical Information and accompanying UNITS below each compound. ____NaOH + ____H 2 SO 4  ____H 2 O + ____Na 2 SO 4 2 1 2 1 0.102 M ? mL 35.0 mL Since 1 L = 1000 mL, we can use this to save on the number of conversions Our Goal

33. 32 Determine how many mL of 0.102 M NaOH solution is needed to neutralize 35.0 mL of 0.125 M H 2 SO 4 solution. Now let’s get to work converting. ____NaOH + ____H 2 SO 4  ____H 2 O + ____Na 2 SO 4 2 1 2 1 0.102 M ? mL 35.0 mL = mL NaOH H 2 SO 4 35.0 mL H 2 SO 4 0.125 mol 1000 mL H 2 SO 4 NaOH 2 mol 1 mol H 2 SO 4 1000 mL NaOH 0.102 mol NaOH 85.8 Units Match Solution Stoichiometry: shortcut

34. 33 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 Solution Stoichiometry

35. 34 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)  2H 2 O(l) + BaCl 2 0.40 M 47.1 mL 0.75 M ? mL = mL HCl Ba(OH) 2 47.1 mL 1 mol Ba(OH) 2 HCl 2 mol 0.40 mol HCl HCl 1000 mL 176 Units match Solution Stoichiometry

36. 35 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)? First write a balanced chemical reaction. ____HCl(aq) + ____Ba(OH) 2 (aq)  ____H 2 O(l) + ____BaCl 2 (aq) 2 1 2 1 23.28 mL 0.135 mol L 25.00 mL ? mol L

37. 36 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)? ____HCl(aq) + ____Ba(OH) 2 (aq)  ____H 2 O(l) + ____BaCl 2 (aq) 2 1 23.28 mL 0.135 mol L 25.00 mL ? mol L = mol Ba(OH) 2 L Ba(OH) 2 25.00 x 10 -3 L Ba(OH) 2 Units Already Match on Bottom! 0.0629 Units match on top!

38. 37 48.0 mL of Ca(OH) 2 solution was titrated with 19.2 mL of 0.385 M HNO 3. Determine the molarity of the Ca(OH) 2 solution. We must first write a balanced equation. Solution Stochiometry Problem:

39. 38 48.0 mL of Ca(OH) 2 solution was titrated with 19.2 mL of 0.385 M HNO 3. Determine the molarity of the Ca(OH) 2 solution. Ca(OH) 2 (aq) + HNO 3 (aq)  H 2 O(l) + Ca(NO 3 ) 2 (aq) 2 2 48.0 mL19.2 mL 0.385 M = mol (Ca(OH) 2) L (Ca(OH) 2 ) 19.2 mL HNO 3 48.0 x 10 -3 L ? M units match! 0.0770 Solution Stochiometry Problem: