Chemistry Week 7
Chemistry Week 7 CH 1213
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This 7 page Class Notes was uploaded by Katerina Kushla on Thursday September 29, 2016. The Class Notes belongs to CH 1213 at Mississippi State University taught by Dr. Eric Van Dornshuld in Fall 2016. Since its upload, it has received 5 views. For similar materials see Chemistry 1 in Chemistry at Mississippi State University.
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Date Created: 09/29/16
Chapter 4: Week 7 9/27-29/16 Chapter 4: Stoichiometry of Chemical Reactions Part 3: Reaction Classification and Stoichiometry and Quantitative Analysis Chemical Reaction Names Combination Reaction – two reactants form one product NH (3) + HCl(g) NH Cl(s) 4 Decomposition Reaction – a reactant decomposes into two or more products CaCO (s) CaO(s) + CO (g) 3 2 Combustion Reaction – a reaction where a substance containing C, H, and (sometimes) O is burned in the presence of oxygen to produce only CO 2nd H O2 CH O2l) + O (g) 2 CO (g) + H 2(l) 2 Disproportionation Reaction – a redox reaction that consists of an element that undergoes oxidation and reduction 2H O2(a2) 2H O(l) +2O (g) 2 4.3 Reaction Stoichiometry Stoichiometry – relationships between amounts of reactant and products in a chemical reaction 4Fe(s) + 3O (g) 2Fe O (s) 2 2 3 So to make 6 moles of Fe2O3, 12 moles of solid iron would be required. If you had 9 moles of 3O2and an unlimited amount of Fe, you could make 6 moles of Fe2O 3 Chapter 4: Week 7 9/27-29/16 What mass (in kg) of Fe is required to react with 5.00 kg of O ? 2 Summary of Steps: 1. Balance the chemical equation 2. Convert mass to moles 3. Implement appropriate stoichiometric relationship from equation 4. Convert moles to mass 4.4 – Reaction Yields Limiting Reactant – the reactant that limits the maximum number of product you can make; it is the substance that is to be completely consumed first; once a reaction is completely consumed the reaction stops. Excess Reactant – any reactant that is not the limiting reactant; will be present after a reaction stops Theoretical Yield – the amount of product that may be produced by a reaction under specific conditions as calculated per the stoichiometry of a balanced chemical equation Actual Yield – the amount of product actually (less that the theoretical yield) Percent Yield – the extent to which a reaction’s theoretical yield is achieved; unit-less quantity if actual and theoretical yields are expressed using the same units Chapter 4: Week 7 9/27-29/16 Limiting Reactant Example If you had 2.00 g Fe and 1.55 g O ,2how much product could you make (in g) according to the following chemical equation? How much excess reactant (in g) would be left over? Limiting Reactant (Fe) Excess Reactant (0.672g O ) Fe O (2.87g) 2 2 3 Chapter 4: Week 7 9/27-29/16 Summary of Steps: 1. Convert mass to moles 2. Determine product yields; identify limiting reactant 3. Calculate product mass 4. Find how much excess reactant is consumed and leftover 5. Convert mass to moles Percent Yield Example 3.45 g Fe and excess O 2ere reacted and produced 4.12 g of Fe O 2 3hat is the percent yield? Actual Yield (4.12 g) Theoretical Yield (4.95 g) Percent Yield (83.23%) Summary of Steps: 1. Convert mass to moles 2. Determine theoretical yield (in g); use stoichiometric coefficients 3. Find percent yield Chapter 4: Week 7 9/27-29/16 4.5 – Quantitative Chemical Analysis Quantitative Analysis – the determination of the amount or concentration of a substance in a sample Titration Analysis – quantitative chemical analysis method that involves measuring the volume of a reactant solution required to completely react with the analyte in a sample A buret is filled with a solution with a known concentration of a substance (titrant). The solution is incrementally added to a flask containing a solution with an unknown concentration of some substance (analyte). Once a complete reaction has occurred (equivalence point), the volume of added titrant is recorded (end point). The concentration of the analyte is then calculated. Chapter 4: Week 7 9/27-29/16 Titration Example Titration of a 20.0 mL sample of acid rain required 1.7 mL of 0.0811 M NaOH to reach the end point. The acidity of the rain is due to the presence of sulfuric acid. What was the concentration of the sulfuric acid in this sample of rain? Summary of Steps: 1. Convert volume of NaOH to mmol NaOH 2. Determine mmol H SO f2om 4toichiometric relationship 3. Calculate concentration of H SO2(mo4arity equation) Chapter 4: Week 7 9/27-29/16 Gravimetric Analysis – a technique where a sample is subjected to some treatment that causes a change in the physical state of the analyte that permits its separation from other components in the sample Gravimetric Analysis Example A sample of gallium bromide weighing 0.165 was dissolved in water and treated with silver nitrate resulting in the precipitation of 0.299 g AgBr. Find the percent of Ga (by mass) in the GaBr 3ample. Summary of Steps: 1. Write out a balanced chemical equation 2. Convert mass of AgBr to moles 3. Find moles of GaBr f3om stoichiometric relationship 4. Convert moles to mass 5. Find mass percent
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