 # Section 1.3.  Limiting reactant  Excess reactant  Yields:  Theoretical yield  Percentage yield  Experimental yield  Avogadro’s law  Molar volume.

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Section 1.3

 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

 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

 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

 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?

 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

 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?

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?

 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?

 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.

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?

 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

 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

 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

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

__________  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

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

___________  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.

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

 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.

P 1 V 1 = P 2 V 2

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

 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

 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

 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)

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

 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?

 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

 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 ?

 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.

 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

 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

 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

 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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