STOICHIOMETRY – MOLE RELATIONSHIPS IN CHEMICAL REACTIONS 1 Experiment: Mole Relationships in Chemical Reactions (pg. 235) Purpose: To investigate the relationship between the moles of products and reactants in a chemical reaction. Then compare the relationships with the balanced chemical equation. The balanced chemical reaction for NaHCO3 reacting with HCl is: Procedure: Follow the procedure steps on page 235, with the following changes: When you add the NaHCO3 to the microplate, put a small scoop (not 1g). Use 10 M HCl instead of 8M HCl. When you’re adding the HCl to the NaHCO3, be careful not to add too much at once. Add it drop-by-drop, so that it doesn’t overflow (if it does overflow, that will obviously affect your mass measurements). You can use the dropper of HCl to stir the mixture to ensure all the NaHCO3 reacts. Observations: Mass of NaHCO3 Mass of microplate and NaHCO3 Mass of microplate, NaHCO3 and HCl before reaction Mass of microplate, NaHCO3 and HCl after reaction is done Calculate mass difference (which is mass of CO2 produced) Analysis: Answer the analysis questions pg. 235 #1-6. Record your answers here. STOICHIOMETRY – MOLE RELATIONSHIPS IN CHEMICAL REACTIONS 2 Mole Calculations using the Balanced Chemical Equation Reference: 7.1 pages 243 - 249 Stoichiometry is the study of the relative quantities of reactants and products in chemical reactions. Based on the results of moles relationship experiment, what is the relationship between moles of reactants and products compared to the balanced chemical equation? Using this information we can take it a step further and use the mass of any reactant or product to predict he mass of another reactant or product using the following process: Mass of Moles of Use mole ratio in Moles of Mass of known known chemical equation to unknown unknown material material determine number of material material moles of unknown material Sample problem – Mole Ratios Solving problems using the mole ratios given by the coefficients in the reaction follow a few simple steps: 1. Write a balanced chemical equation for the reaction (if not already given). 2. Identify the known moles and the unknown moles. 3. Identify the ratio of the unknown to known moles given by the equation. Example 1: Vanadium reacts with oxygen to form divanadium pentoxide. Determine the number of moles of vanadium needed to produce 7.47 mol of divanadium pentoxide. 4V + 5O2 2V2O5 4V + 5O2 V2O5 ______ mol V 7.47mol V2O5 * notice how the moles of known units cancel out in this calculation!* STOICHIOMETRY – MOLE RELATIONSHIPS IN CHEMICAL REACTIONS 3 Example 2: Now you try it! If 2.0 mol of iron (steel wool) are reacted with enough copper(I) chloride, how many moles of copper metal should be produced? Fe + CuCl FeCl2 + Cu Practice Questions pg 238 # 4, 5, 6, 7; pg 240 #9, 10 STOICHIOMETRY – MOLE RELATIONSHIPS IN CHEMICAL REACTIONS 4 Sample problem – Mass to Mass and Mass to Particle Calculations Solving this type of problem involves five steps that should be carried out in the correct sequence: 1. Write a balanced chemical equation. 2. Read the question and identify the known and unknown materials in the equation. 3. Calculate the number of moles of the known or given material. 4. Use the mole ratio from the balanced chemical equation to determine the number of moles of unknown material. 5. Convert the number of moles of unknown material to whatever is required in the question (mass or molecules). Key Point: To use the ratio of reactants and products in the balanced equation, you MUST HAVE THE quantities in __________________. So if the question gives you mass, you must convert to moles first. Example 2: Calculate the mass of oxygen produced by the decomposition of 12.26g of potassium chlorate into potassium chloride and oxygen. 2KClO3 2 KCl + 3O2 2KClO3 2 KCl + 3O2 ________ mol KClO3 ______mol O2 12.26 g KClO3 ? g O2 STOICHIOMETRY – MOLE RELATIONSHIPS IN CHEMICAL REACTIONS 5 Example 3: Now You Try It! What mass of sodium would have to be burned in an excess of oxygen to produce 6.2g of sodium oxide? 4Na + O2 2Na2O 4Na + O2 2Na2O ? g Na 6.2 g Na2O Example 4: Nitrogen gas and sodium metal are generated by the decomposition of sodium azide, NaN3. How many atoms of Na are produced when 80.0g of N2 are generated in this reaction? 2NaN3 3N2 + 2Na 2NaN3 3N2 + 2Na ` 80.0g N2 ? atoms Na Exercise: 1. Questions 11, 12, 13, 14; page 244 2. Questions 16, 17, 18; page 246 3. Questions 20, 21, 22; page 248-249 STOICHIOMETRY – MOLE RELATIONSHIPS IN CHEMICAL REACTIONS 6 Experiment: Limiting Reactant in a Chemical Reaction Complete Thoughtlab: The Limiting Item (page 252) and/or Investigation 7-A (page 255). Calculations Involving the Limiting Reactant Reference: Section 7.2 pages 251-257 These problems are only slightly different from mass-mass problems. The key difference is that information will be given for and you will always be asked for the amount of . Example: Zinc and sulfur react to form zinc sulphide. If 6.00g of zinc and 4.00g of sulfur are available for reaction, determine the limiting reactant and the mass of zinc sulfide produced 8Zn + S8 8ZnS 8Zn + S8 8ZnS ______ mol Zn ______ mol ZnS OR ______ mol S8 ______ mol ZnS 6.00 g Zn 4.00g S ______g ZnS Practice: 1. Questions 24 page 254. 2. Questions 27, 28, 30; page 258. STOICHIOMETRY – MOLE RELATIONSHIPS IN CHEMICAL REACTIONS 7 Percentage Yield Reference: Section 7.3 pages 260 – 270 When calculating the mass of products using the process given, it is a theoretical yield that is being calculated. The theoretical yield is the amount that would be obtained if everything in the reaction (experiment) was perfect. Both in the school laboratory and in industry things are not 100% perfect in any chemical production. This is the number you have been calculating in the mass-mass and limiting reactant problems. The amount of product that you actually obtain in an experiment is called the actual yield. To the percentage yield from an experiment, you divide the actual yield by the theoretical yield and multiply by 100%. Example: The actual yield of an experiment is 12.6 g when the calculated value of the theoretical yield was 15.0g. The percentage yield would be calculated as follows: percentage yield = actual yield x 100 = 12.6 g x 100 = 84.0% yield theoretical yield 15.0 g Exercise: 1. Question 32, 33 page 262. 2. Questions 35, 27 page 264. UNIT REVIEW Chapter 5 Terms: average atomic mass, isotopic abundance, the mole, Avogadro’s number, molar mass (of an element or a compound), the mole equation, the mole/mass equation Pg. 170 # 5, pg. 193 # 9, 13(a), pg. 194 # 14(a), 16(d), 19, 25 Chapter 6 Terms: Law of Definite Proportions, Mass Percent, Percentage composition of a compound, Simplest formula vs. Molecular formula, Hydrates and Anhydrous compounds Pg. 229 # 6 (a) and (b), pg. 211, #14 and 16, pg 218 # 20, pg 230 # 18 (a)(b), pg 225 #23 to 25; Plus Problem Set Chapter 7 Terms: mole ratios, coefficients, stoichiometry, limiting reactant, excess reactant, actual yield, theoretical yield, percentage yield Pg. 271 #1, 3, 5, 7, 9(a)(b), 15(a)(b) *Pay attention to SIG DIGS and UNITS in your review, because they will count on the test!
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