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Section 1.3

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Limiting reactant Excess reactant Yields: Theoretical yield Percentage yield Experimental yield Avogadro’s law Molar volume Ideal gas Charles’ Law Boyle’s Law Universal gas constant Solutions Solute Solvent Concentration Titrations Equivalence point

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5 major types: Describe each one Synthesis 2Na + Cl 2 2NaCl Single replacement 3CaBr 2 + N 2 Ca 3 N 2 + 3Br 2 Double replacement HCl + NaOH NaCl + HOH Combustion CH 4 + 2O 2 CO 2 + 2H 2 O Decomposition CaCO 3 CO 2 + CaO

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Theoretical yield: the yield you are expecting to get. Determined through calculation Experimental yield: the yield you obtained in the experiment. May obtain more or less than expected Percentage yield: the comparison of the theoretical with the experimental yields. Experimental/theoretical * 100 = % yield Very important for: Data analysis Determining efficiency of method

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Limiting reactant: this reactant determines the amount of product that can be produced Excess reactant: this reactant has a larger quantity than is needed for the reaction Ex: 2H 2 + O 2 2H 2 O If have 2 mol H 2 and 2 mol O 2, which is limiting and which is excess? How can you tell? How much H 2 O can be made?

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Use the equation: 2NH 3 + 3CuO N 2 + 3Cu + 3H 2 O If given 20.5 g NH 3 and 100.3 g CuO, which reactant is in excess? Limiting? Determine the number of moles for each reactant Determine the mole ratio for the reactants from the equation: 2/3 = 0.667 Determine the mole ratio for the reactants from the moles calculated: 1.203/1.261 = 0.95 Since 0.95 (given) > 0.667 (needed), then, NH 3 is in excess and CuO is limiting

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Given the previous problem, how can we determine the amount of product that could be formed from each reactant? Follow a path from reactant to product for each given reactant: 2NH 3 + 3CuO N 2 + 3Cu + 3H 2 O If given 20.5 g NH 3 and 100.3 g CuO, how much N 2, in grams, can be formed?

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2NH 3 + 3CuO N 2 + 3Cu + 3H 2 O If given 15.5 g NH 3 and 25.3 g CuO, how much Cu, in grams, can be formed? If given 25.4 g NH 3 and 20.7 g CuO, how much H 2 O, in grams, can be formed?

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If given that 9.3 g of N 2 was obtained in the previous experiment, what is the percentage yield? Experimental/theoretical * 100 9.3/____ *100 = If given that 50 g of Cu was obtained in the previous experiment, what is the percentage yield?

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Equal volumes of all gases, when measured at the same temperature and pressure, contain an equal number of particles Ex: At the same temp and pressure, there is 1 mol of H 2 in the flask, there is the same volume of O 2 in a second flask. How many moles are there of O 2 ? Because the volume, pressure and temp are the same, there is 1 mol of O 2 in the second flask as well.

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2CO + O 2 2CO 2 At equal Pressure and Temp, If 20 cm 3 of CO react with 20 cm 3 of oxygen, what volume of CO 2 is produced? Which reactant is limiting? Excess?

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Describes the behavior of an ideal gas Particles are hard spheres with insignificant volume, mostly empty space, no attractive or repulsive forces Particles in constant, rapid motion All collisions are perfectly elastic

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PRESSURE – force of gas collisions per unit area of the container (kPa or Pa) TEMPERATURE – average kinetic energy of the gas particles (KELVIN) VOLUME – space the gas takes up (dm 3 ) Standard Temperature and Pressure (STP): Temperature: 0 C or 273.15 K Pressure: 100 kPa or 1 atm or 760 mmHg

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Convert the following to Kelvin: 25°C-30°C 0°C100°C Convert the following to kPa: 4 atm850 mmHg 0.68 atm635 mmHg Convert the following to dm 3 : 3 m 3 4 cm 3 0.7 m 3 0.8 cm 3

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Inverse relationship between pressure and volume What is constant? P 1 V 1 = P 2 V 2

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__________ VOLUME-as volume decreases, same number of particles collide with sides of a smaller container, therefore pressure __________ (temperature is constant) INCREASES DOUBLE INVERSE Half the volume… ________ the pressure ________ the pressure This is an ________ This is an ________ relationship

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Pressure and temperature of a gas are directly proportional Temperature must be in KELVIN

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___________ TEMPERATURE- as temperature increases, particle velocity is faster, increasing collisions, therefore pressure ___________ (volume constant) INCREASES DOUBLE DIRECT Double the Kelvin temperature… Double the Kelvin temperature… __________ the pressure __________ the pressure This is a _________ This is a _________ relationship.

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Direct relationship between temperature and volume Temperature must be in KELVIN What is constant?

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As temperature of a gas decreases, the particle velocity is slower, making __________ collisions, and the volume of the gas _______________. FEWER DIRECT HALF DECREASES Half the Kelvin temperature, ___________ the volume. This is a __________ relationship.

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P 1 V 1 = P 2 V 2

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COMBINED GAS LAW Combination of three prior laws Must remember formula Identify as having BEFORE AND AFTER CONDITIONS

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Initially a gas with a pressure of 84 kPa and a temperature of 35ºC is heated an additional 230 degrees, what will the new pressure be? Assume the volume of the container is constant. A car has an internal volume of 2600 liters. If the sun heats the car from a temperature of 20ºC to a temperature of 55ºC, what will the pressure inside the car be? Assume the pressure was initially 760 mm Hg

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AMOUNT OF GAS- (temperature is constant) As the number of gas particles increases, the number of collisions _______, therefore the pressure_______ Double Double the number of particles…… the pressure INCREASE INCREASES DOUBLE

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Considering all gas molecules to be the same, can use IDEAL GAS LAW Identity as having only one (1) set of conditions the number of moles (or a mass that has to be converted)

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PV = n RT Pressure kPa (or Pa ÷1000) Volumedm 3 Temperature Kelvin n is number of MOLES R is Universal gas constant 8.314

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If four moles of a gas at a pressure of 5.4 atmospheres have a volume of 120 dm 3, what is the temperature? How many moles of gas are contained in 890.0 cm 3 at 21.0°C and 750.0 mm Hg pressure?

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Difference between a liquid and a solution? Does a solution have to be in liquid form? Concentration, c, depends on Moles, n, of solute Volume, V, of overall solution Therefore: mol/dm 3 are your typical units Dilutions lower the concentration of solutions This comes with changes in concentration c 1 v 1 =c 2 v 2 The number of moles is constant so cv=n for each

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How would you prepare 50cm 3 of a 0.20 mol dm -3 solution of NaCl? What is the concentration of a 1.5 mol dm -3 solution of sodium sulfate in g dm -3 ?

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Determine the final concentration of a 50 cm 3 solution of HCl of concentration 0.25 mol dm -3, which is diluted to a volume of 250 cm 3. What is the initial concentration of a solution if 50 cm 3 is added to a solution with final volume 300 cm 3 and concentration of 0.1 mol dm -3.

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Titration used to determine reacting volumes precisely Determining the unknown concentration of an known solution First solution of unknown concentration in flask with indicator Second solution of known concentration in pipette or burette Drops of second solution into first (count volume) When equivalence point is reached, know both volumes of reaction (end-point) Solve for moles of unknown concentration For example, finding the concentration of a bottle of NaOH

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15.0 cm3 of 0.2 mol dm-3 sodium hydroxide solution is titrated with dilute hydrochloric acid. If 7.6 cm3 is needed to neutralize the solution, what was the concentration of the initial acid? 1- write balanced equation 2- calculate moles of base (solution with concentration given) 3- do mole ratio conversion 4- divide by the volume to determine concentration

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This is used when the titration point has passed End-point is hard to identify One reactant is impure How to do back titration: A known excess of one reactant is add to mixture Unreacted excess is determined by titration against a standard solution (known). Subtract the amount of unreacted reactant from original amount used This will result in determining reacting amount

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1.435 g sample of dry CaCO 3 and CaCl 2 mixture was dissolved in 25.00 dm 3 of 0.9892 mol dm -3 HCl solution. What was CaCl 2 percentage in original sample, if 21.48 dm 3 of 0.09312 mol dm -3 NaOH was used to titrate excess HCl? CaCO 3 + 2HCl CaCl 2 + CO 2 + H 2 O (H 2 CO 3 ) During titration 21.48 dm 3 × 0.09312 mol dm -3 = 2.000 mole HCl was neutralized. Initially there was 25.00 dm 3 × 0.9892 mol dm -3 = 24.73 mole of HCl used, So during CaCO 3 dissolution (dissolves), 24.73 mol - 2.000 mol = 22.73 mole of acid reacted. Mole ratio of 1CaCO 3 :2HCl Original sample contained 22.73 mol/2mol = 11.37 mol of CaCO 3, or 1.137 g (assuming molar mass of CaCO 3 is 100.0 g) So original sample contained 1.137g/1.435g×100%=79.27% CaCO 3 and 100.0-72.27%=20.73% CaCl 2.

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