1. Stoichiometry Chemistry 6.0

2. The Mathematics of Chemical Reactions: STOICHIOMETRY I. Balanced Chemical Equations A. Provide qualitative and quantitative information Conservation of Matter B. Supports the Law of Conservation of Matter 2H 2 + O 2  2H 2 O The above equation is interpreted in terms of particles as follows: 1. 2 molecules of H 2 react with 1 molecule of O 2 to produce 2 molecules of water. The ratio of H 2 to O 2 is 2:1. or 2 moles of H 2 react with 1 mole of O 2 to produce 2 moles of water. The ratio of H 2 to O 2 is 2:1.

3. 2. It is more convenient to interpret the coefficients as number of moles, because we measure amounts of substances by massing. C. Stoichiometry 1. The study of the quantitative relationships that exist in a formula or a chemical reaction. 2. Importance a. Provides for the safe, economical and reproducible manufacture of chemicals. b. Provides for the safe administration of pharmaceuticals. D. Proof of the conservation of matter in a balanced equation 1. Convert all reactants and products to their mass equivalents. 2. Sum up the mass of reactants and compare the sum of the mass of products.

4. II. Stoichiometry Problems A. Steps to Solve Problems 1.Write a balanced equation. 2.Identify the given (  ) and the unknown or required substance (?). 3.Convert mass of given into moles. 4.Use the mole (molar) ratio to convert from given to required substance. 5.If needed, convert moles of required into mass of required substance.

5. B. Examples 1. How many moles of oxygen are required to react with 16 moles of hydrogen in the production of water? H 2 + O 2  H 2 O 2 2  ? 1 mole O 2 2 moles H 2 = 8.0 moles O 2 Mole ratio links 1 substance to another in a reaction. Required in problem solving. 16 moles H 2

6. 2.Antimony reacts with water to produce antimony(III) oxide and hydrogen. How many moles of hydrogen are produced from 7.5 moles of antimony? Sb + H 2 O  Sb 2 O 3 + H 2 233 11 mol H 2

7. 3.What mass of aluminum oxide can be prepared by the reaction of 67.5 g of aluminum in a synthesis reaction? Al + O 2  Al 2 O 3 432 128 mol Al 2 O 3

8. 4.Sodium bicarbonate, a.k.a. baking soda, can be used to extinguish a fire. When heated, it decomposes to give carbon dioxide gas which smothers the fire. It also produces sodium carbonate and water. If a sample contains 4.0 g of sodium bicarbonate, what mass of carbon dioxide is produced? 2 NaHCO 3  Na 2 CO 3 + H 2 O + CO 2

9. III. Percent Yield A. Expected Yield: the amount of product that should be produced (theoretical) B. Actual Yield: the amount of product that is actually produced (experimental) C. Percent Yield: percent of expected yield that was obtained % Yield = (actual yield/expected yield) x 100

10. D. Steps to Solving Percent Yield Problems 1.Write a balanced equation 2.Identify the given (  ) which is the mass of reactant, and identify the actual yield. 3.Solve for the expected mass of product using the given mass of reactant. 4.Calculate the % yield. % Yield = actual yield x 100 expected yield

11. E. Examples 1. A reaction between 2.80 g aluminum nitrate and excess sodium hydroxide produced 0.966 g of aluminum hydroxide in this double replacement reaction. Calculate the % yield. Al(NO 3 ) 3 + 3NaOH  Al(OH) 3 + 3NaNO 3 1.03 g Al(OH) 3

12. 2. Determine the % yield for the reaction between 3.74 g of sodium and excess oxygen if 4.24 g of sodium oxide is recovered in the direct combination reaction. 4Na + O 2  2Na 2 O

13. IV. Limiting Reactants A.Definition: the reactant that determines, or limits, the amount of product(s) formed in a chemical reaction B.Problem Solving Tips 1.The limiting reactant is not necessarily the reactant present in the smallest amount 2.When you are given the amounts of 2 or more reactants, you should suspect that you are dealing with a limiting reactant problem.

14. C. Steps 1.Write a balanced equation 2.Calculate the number of moles of each reactant 3.Compare the mole ratios of the reactants as available ratio (from the given masses) and the required ratio (from the coefficients) 4.Identify the limiting reactant, and use it to calculate the mass of product formed.

15. D. Examples 1. What mass of CO 2 could be formed by the combustion of 16.0 g CH 4 with 48.0 g O 2 ? CH 4 + 2O 2  CO 2 + 2H 2 O 33.0 g CO 2 Oxygen is the limiting reactant

16. 2. What is the maximum mass of nickel(II) hydroxide that could be prepared by mixing 25.9 g nickel(II) chloride with 10.0 g sodium hydroxide? NiCl 2 + 2NaOH  Ni(OH) 2 + 2NaCl 11.6 g Ni(OH) 2 Sodium Hydroxide is the LR

17. VII. Stoichiometry and Heat Changes 1.How much heat is released when 22.0g of propane is burned? C 3 H 8 + 5O 2 → 3CO 2 + 4H 2 O ΔH  = -2.22x10 3 kJ 2.How much carbon dioxide is produced, in grams, when 2,500 kJ of energy is released? -1.11x10 3 kJ (released) 150 g CO 2

18. Hess’s Law Hess’s Law states that if a series of reactions are added together, the enthalpy change of the net reaction will be the sum of the enthalpy changes of the individual steps.

19. Steps for using Hess’s Law Identify the compounds Locate the compounds on the periodic table Write a reaction from the table. Write the appropriate “sub equation.” – If needed, multiply equation and enthalpy change. – If you reverse the reaction, change sign of enthalpy change. Add equations. Add enthalpy changes.

20. The standard enthalpy change of a reaction is equal to the sum of the standard molar enthalpies of formation of the products multiplied by its coefficient, n, in the balanced equation, minus the corresponding sum of standard molar enthalpies of formation of reactants.  H rxn  = ∑ n  H f, products - ∑ n  H f, reactants VIII. Determining ∆H Using Heats of Formation

21. ΔH from ΔHf Problem: Using the Heats of Formation Table, calculate the  H for the following reaction: SF 6 (g) + 3H 2 O(l)  6HF(g) + SO 3 (g) Write the thermochemical equation for this reaction: SF 6 + 3H 2 O + 45.1kJ  6HF + SO 3 45.1 kJ

22. ΔH from ΔHf Problem: Using the Heats of Formation Table, calculate the standard heat of combustion for propane. C 3 H 8 (g) + 5O 2 (g)  3CO 2 (g) + 4H 2 O(g) Write the thermochemical equation for this reaction: C 3 H 8 (g) + 5O 2 (g)  3CO 2 (g) + 4H 2 O(g) + 2043.9kJ -2043.9 kJ