Week 7 Chemistry Notes
Week 7 Chemistry Notes CHEM 1127Q 001
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This 3 page Class Notes was uploaded by Amelia on Wednesday October 12, 2016. The Class Notes belongs to CHEM 1127Q 001 at University of Connecticut taught by Fatma Selampinar (TC), Joseph Depasquale (PI) in Spring 2016. Since its upload, it has received 10 views. For similar materials see General Chemistry in Chemistry at University of Connecticut.
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Date Created: 10/12/16
10/10 Chapter 4- Stoichiometry and Chemical Reactions (Cont.) Stoichiometry 1) Molar ratio of reactants and products to each other 2) Reactant stoichiometry when burning propane is 1 propane to 5 oxygen molecules 3) Propane is equal to 3 CO after burning 2 4) You need to know the number of moles/molecules for this to work. Anything else is not applicable. a) For example, if you are given mass, you need to turn that into moles by dividing by Molar Mass, and then you find the moles of the other given element and turn that back into mass by multiplying by Molar Mass. b) Example i) C H (g) + 5O (g) → 3CO (g) + 4H O(l) If I burn 2.00g of propane how much 3 8 2 2 2 water do I create in mL. ii) C3 83(12.001g/mol) + 8(1.008g/mol) = 44.094g/mol C H 3 8 iii) 2.00g ÷ 44.094g/mol C H = 0.0453m8es C H 3 8 iv) 0.0454 moles C H ×(4H O ÷ 1 C H ) = 0.181 moles H O 3 8 2 3 8 2 v) 0.181 moles H O ×18g/mole = 3.266g/mole H O 2 2 vi) 3.266g/mole H O÷1 g/mL = 3.266mL H O 2 2 vii) 3.27mL H O 2 5) Important Information a) Mass = MM × n b) M = moles/liter or [X] c) You can't destroy or create matter. Balance equations d) When converting between molecules always convert through moles 6) Titrations a) Acid react with base + - b) Always H + OH → H O 2 c) Measuring the volume of a Standard Solution required to react with a measured amount of sample. d) Equivalence Point: point at which the reaction is complete in a titration. e) Example i) During a titration of 500mL of an acid solution of unknown concentration you reach the equivalence point after adding 25mL of 1M Calcium Hydroxide. What + is [H ] in the unknown solution? ii) H (?) + OH (25mL and 1M) → H O 2 iii) 0.025L ×1 mole/L Ca(OH) = 0.025 moles Ca(OH) 2 2 iv) 0.025 moles Ca(OH) ×(2OH ÷2Ca(OH) ) = 0.05 moles OH 2 - - + - + v) 0.05 moles OH ×(1H ÷ 1 OH) = 0.05moles H + + vi) 0.05moles H ÷0.500L = 0.1 mole/L H f) Example 2 i) You have 500mL of Sulfuric Acid but you don’t know the concentration. You titrate the solution with 25mL of 2M NaOH but overshoot the equivalence point. You then went back titrate with 3 mL of 0.1M HCL. What is the concentration of the sulfuric acid solution? ii) H + OH → H O 2 iii) H 2 + 4L = NaOH - iv) 0.025L ×2 moles/L NaOH = 0.05moles NAOH × (1 OH ÷ 1 NaOH) = 0.05 - moles OH + v) 0.003L ×0.1 moles/L HCL = 0.0003 moles HCL ×(1H ÷ 1 HCL) = + 0.0003moles H vi) 0.05 moles OH ×(1H ÷ 1OH) = 0.05 moles H total - 0.0003 moles H HCL = + 0.0497 moles H H SO + 2 4 vii) H 2 → 4 + SO + 4- + + viii) 0.0497 moles H × (1 H SO ÷ 2 H )2 0.0485 moles 1 H SO ÷ 0.500L = 2 4 0.0497M H SO (n2 acc4ate sig figs) g) Example 3 i) Consider a 1.50g mixture of magnesium nitrate and magnesium chloride. After dissolving this mixture in water, 0.500M silver nitrate is added dropwise until the precipitation is complete. The mass of the precipitate formed is 0.641g. Calculate the mass percent of magnesium chloride in the mixture. (1) AgNO (aq) +3g(NO ) → does not 3 2, no precipitate (2) 2AgNO (aq) + MgCL → 2AgCl(s) + Mg + 2NO 2+ - 3 2 3 (3) AgCl(s) = 107.877g/mole + 35.45g/mole = 143.32g/mole (4) 0.641 ÷143.32g/mole AgCl = 0.00447moles AgCl(s) (5) 0.00447 moles AgCl × (1 MgCl ÷2 AgCl) = 0.0022352oles MgCl 2 (6) 0.002235 moles MgCl × 95.21g/mole 2Cl = 0.2128g MgCl 2 2 (7) Mass % = (Mass MgCl ÷ total mass)×12 (8) % = (0.2128g MgCl ÷ 1.50g) ×102= 14.2% ii) What is the minimum volume of silver nitrate that must have been added to ensure the complete formation of the precipitate? 2+ - (1) 2AgNO (aq) + 3CL → 2AgCl(s) + M2 + 2NO 3 (2) 0.00447moles AgCl(s)×(2AgNO ÷ 2AgCl) = 0.00447moles of AgNO 3 3 (3) 0.00447moles of AgNO ÷0.500moles/L (M) = 0.00894L AgNO 3 3 (4) 8.94mL of AgNO 3 10/12 7) Chemical Reactions a) Balancing reactions is great. You should always do it. b) But in real life we don’t always react 1 mole of to 3 moles of something else (exact ratios). c) Sometimes we carry out reactions where there is an excess of one or the other i) C 3 +80O → 3CO2+ 4H O + 52 2 2 ii) The one that is not in access is called the limiting element. 8) Limiting Reactants a) Write out the balanced equation b) Turn everything into moles if it isn’t already. c) Check which reactant can make more product. d) Example i) 3g of Magnesium metal plus 1 mole of carbon dioxide react to make carbon and Magnesium Oxide. What is the limiting reactant? ii) Mg(s) + CO (s) → C(s) + MgO(s) 2 iii) 2Mg(s) + CO (s) 2C(s) + 2MgO(s) iv) .1 moles of Mg (approximately) × 0.5 Carbons per Mg = 0.05 moles C v) 1 mole CO × 12÷1CO = 1 mole C 2 vi) Magnesium runs out first so it’s the limiting reactant. We can make 0.05 moles of Carbon according to the theoretical yield. e) Reactions don’t always work perfectly. i) After we reacted the Magnesium with carbon dioxide we got 0.05g of Carbon. How does that compare to the calculated yield? (1) 0.05g C ÷ 12.01 g/mole C = 0.004 moles C (2) This is the experimental yield. f) Percent Yield = Experimental/Theoretical ×100 i) 8% = 0.004 moles C / 0.05 moles C × 100 g) Another Example i) You are given a 1cm piece of sodium metal (0.968 g/cm ) and pace it in 25 mL water. What reactant is in excess? (1) 2Na(s) + 2H O → H (g) + 2NaOH 3 2 2 3 (2) 1 cm Na ×0.968 g/cm = 0.968g Na ÷ 22.99g/mole = 0.0421 moles Na (3) 25 mL H O ×12mL = 25g H O ÷ 18 g/mole2 1.389 moles H O 2 (4) 0.0421 moles Na × (1 H ÷ 2 Na) = 0.02105 moles H 2 (5) 1.389 moles H O ×(12 ÷ 2 H O) = 0.6945 2les H 2 (6) H O2s in excess, Sodium is limiting ii) How much of the excess reactant remains (in grams) after the reaction is completed? (1) 0.0421 moles Na × (2 H O ÷ 2 Na) = 0.0421moles H O consumed 2 2 (2) 0.0421moles H O cons2ed × 18 g/mole = 0.7578g H O consumed 2 (3) 25g - 0.7578g = 24.24g left over iii) How many grams of H were generated? 2 (1) 0.0421 moles Na × (1 H ÷ 2 Na) = 0.02105 moles H × 2 g/mole = 2 0.0421g H 2 h) Important concepts i) Limiting reactants ii) Theoretical and Experimental Yields iii) Percent Yield
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