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Section 12.2 Chemical Calculations

Chapter 12 Stoichiometry. Section 12.2 Chemical Calculations. Mole Ratios. These mole ratios can be used to calculate the moles of one chemical from the given amount of a different chemical .

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Section 12.2 Chemical Calculations

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  1. Chapter 12 Stoichiometry Section 12.2 Chemical Calculations

  2. Mole Ratios These mole ratios can be used to calculate the moles of one chemical from the given amount of a different chemical Example: How many moles of chlorine are needed to react with 5 moles of sodium (without any sodium left over)? Step 1:Write the chemical equation. Step 2:Balance the equation. 2 Na + Cl2 NaCl 2 Step 3: (Write the mole ratio) 2 mol 1 mol (Write the given) Step 4: 5 mol x Step 5:

  3. Mole - Mole Conversions Example: • How many moles of sodium chloride will be produced if you react 2.6 moles of chlorine gas with an excess (more than you need) of sodium metal? 2 Na + Cl2 2NaCl 1 mol 2 mol 2.6 mol x

  4. Mole - Mole Conversions Example: • How many moles of O2 are produced when 3.34 moles of Al2O3 decompose? 2 4 Al2O3® Al+O2 3 2 mol 3 mol 3.34 mol x

  5. Mole - Mass Conversions • Most of the time in chemistry, the amounts • are given in grams instead of moles • We still go through moles and use the • moleratio, but now we also use molar • mass to get to grams

  6. Example: How many grams of chlorine are required to react completely with 5.00 moles of sodium to produce sodium chloride? 2 Na + Cl2 NaCl 2 2 mol 1 mol 5 mol x Change moles to grams: No. of moles X molar mass 2.5 X 71 = 178 g Cl2

  7. Example: • Calculate the mass in grams of Iodine required to react completely with 0.50 moles of aluminum. 3 2 I2 + Al ® AlI3 2 3 mol 2 mol x 0.5 mol Change moles to grams: No. of moles X molar mass 0.75 X 254 = 191 g I2

  8. Mass – Mole Conversions • We can also start with mass and convert to • moles of product or another reactant • We use molar mass and the mole ratio to get • to moles of the compound of interest

  9. Example: • Calculate the number of moles of ethane (C2H6) needed to produce 10.0 g of water • 2 C2H6 + 7 O2 4 CO2 + 6 H2O 2 mol 6 mol x 10 g But before making cross multiplication, change 10 grams to moles 10 / 18.02 = 0.555 mol

  10. Mass – Mass Conversion Example: • Calculate how many grams of ammonia are produced when you react 2.00g of nitrogen with excess hydrogen. N2 + H2  NH3 2 3 1 mol 2 mol 2 gx But before making cross multiplication, change 2grams to moles 2 / 28 = 0.0714 mol Change moles to grams: No. of moles X molar mass 0.143 X 17.03 = 2.44 g NH3

  11. Example: 4 Al+O2 ® 3 Al2O3 2 • Calculate how many grams of oxygen are required to make 10.0 g of aluminum oxide 3 mol 2 mol x 10 g But before making cross multiplication, change 10 grams to moles 10 / 102 = 0.098 mol Change moles to grams: No. of moles X molar mass 0.147 X 32 = 4.7 g O2

  12. Example: (9.6 mol) 2 C2H2 + 5 O2® 4 CO2 + 2 H2O • If 3.84 moles of C2H2are burned, how many moles of O2 are needed? 2 mol 5 mol 3.84 molx

  13. Example: • How many moles of C2H2are needed to produce • 8.95 mole of H2O? (8.95 mol) 2 C2H2 + 5 O2® 4 CO2 + 2 H2O 2 mol 2 mol x 8.95 mol

  14. Example: • If 2.47 moles of C2H2are burned, how many moles of CO2 are formed? (4.94 mol) 2 C2H2 + 5 O2® 4 CO2 + 2 H2O 2 mol 4 mol 2.47 molx

  15. Mass – Mass Problem: 6.50 grams of aluminum reacts with an excess of oxygen. How many grams of aluminum oxide are formed? 4 Al + 3 O2 2Al2O3 4 mol 2 mol 6.5 gx But before making cross multiplication, change 6.5 grams to moles 6.5 / 27 = 0.241 mol Al Change moles to grams: No. of moles X molar mass 0.121 X 102 = 12.3 g Al2O3

  16. Another example: If 10.1 g of Fe are added to a solution of Copper (II) Sulfate, how many grams of solid copper would form? 2 Fe + 3 CuSO4® Fe2(SO4)3 + 3 Cu 2 mol 3 mol 10.1 g x before making cross multiplication, change 10.1 grams to moles 10.1 / 55.8 = 0.181 mol Fe Change moles to grams: No. of moles X molar mass 0.272 X 63.5 = 17.3 g Cu

  17. Volume – Volume Calculations: • How many liters of CH4 at STP are required to completely react with 17.5 L of O2 ? CH4 + 2 O2® CO2 + 2 H2O 1 mol 2 mol x 17.5 L before making cross multiplication, change 17.5 L to moles 17.5 / 22.4 = 0.781 mol O2 Change moles to liters: No. of moles X 22.4 0.391 X 22.4 = 8.76 L CH4

  18. Section 12.3

  19. Limiting Reactant: Cookies 1 cup butter, 1/2 cup white sugar 1 cup packed brown sugar 1 teaspoon vanilla extract 2 eggs 2 1/2 cups all-purpose flour 1 teaspoon baking soda 1 teaspoon salt 2 cups semisweet chocolate chips Makes 3 dozen If we had the specified amount of all ingredients listed, could we make 4 dozen cookies? What if we had 6 eggs and twice as much of everything else, could we make 9 dozen cookies? What if we only had one egg, could we make 3 dozen cookies?

  20. Limiting Reactant • Most of the time in chemistry we have more of one reactant than we need to completely use up other reactant. • That reactant is said to be in excess (there is too much). • The other reactant limits how much product we get. Once it runs out, the reaction s. This is called the limiting reactant.

  21. Limiting Reactant • To find the correct answer, we have to try all of the reactants. We have to calculate how much of a product we can get from eachof the reactants to determine which reactant is the limiting one. • The lower amount of a product is the correct answer. • The reactant that makes the least amount of product is the limiting reactant. Once you determine the limiting reactant, you should ALWAYS start with it! • Be sure to pick a product! You can’t compare to see which is greater and which is lower unless the product is the same!

  22. Example • 10.0g of aluminum reacts with 35.0 grams of chlorine gas to produce aluminum chloride. Which reactant is limiting, which is in excess, and how much product is produced? 2 Al + 3 Cl22 AlCl3 • Start with Al: • Now Cl2: 2 mol 2 mol 10 g x But before making cross multiplication, change 10 grams to moles 10 / 27 = 0.370 mol 0.370 x 133.5 = 49.4 g 3 mol 2 mol 35 g x But before making cross multiplication, change 35 grams to moles 35 / 71 = 0.493 mol 0.329 x 133.5 = 43.9 g

  23. LR Example Continued • We get 49.4 g of aluminum chloride from the given amount of aluminum, but only 43.9 g of aluminum chloride from the given amount of chlorine. Therefore, chlorine is the limiting reactant. Once the 33.6 g of chlorine is used up, the reaction comes to a complete .

  24. Practice • If 10.6 g of copper reacts with 3.83 g sulfur, how many grams of product (cuprous sulfide) will be formed?

  25. 15.0 g of potassium reacts with 15.0 g of iodine. Calculate which reactant is limiting and how much product is made.

  26. Finding the Amount of Excess • By calculating the amount of the excess reactant needed to completely react with the limiting reactant, we can subtract that amount from the given amount to find the amount of excess. • Can we find the amount of excess potassium in the previous problem?

  27. Practice 15.0 g of potassium reacts with 15.0 g of iodine. If 19.6 g of potassium iodide is produced, find the limiting and excess reactants then how much of the excess amount is left over?

  28. Potassium superoxide, KO2, is used in rebreathing gas masks to generate oxygen. How many moles of O2 can be produced from 0.15 mol KO2 and 0.10 mol H2O? Determine the limiting reactant. 4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g)

  29. Limiting Reactant: Recap • You can recognize a limiting reactant problem because there is MORE THAN ONE GIVEN AMOUNT. • Convert ALL of the reactants to the SAME product (pick any product you choose.) • The lowest answer is the correct answer. • The reactant that gave you the lowest answer is the LIMITING REACTANT. • The other reactant(s) are in EXCESS. • To find the amount of excess, subtract the amount used from the given amount. • If you have to find more than one product, be sure to start with the limiting reactant. You don’t have to determine which is the LR over and over again!

  30. Yield • The amount of product made in a chemical reaction. • There are three types • Actual yield- what you get in the lab when the chemicals are mixed • Theoretical yield- what the balanced equation tells you should make. • Percent yield = Actual x 100 % Theoretical

  31. Percent Yield • To determine percentage yield, you need two pieces of information: Theoretical yield = Actual yield = Comes from the stoichiometry Comes from experimental data Actual Yield Theoretical Yield 100% Percent Yield =

  32. Example • 6.78 g of copper is produced when 3.92 g of Al are reacted with excess copper (II) sulfate. • 2Al + 3 CuSO4® Al2(SO4)3 + 3Cu • What is the actual yield? • What is the theoretical yield? • What is the percent yield? • If you had started with 9.73 g of Al, how much copper would you expect?

  33. Details • Percent yield tells us how “efficient” a reaction is. • Percent yield can not be bigger than 100 %.

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