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TOPIC 1 STOICHIOMETRIC RELATIONSHIPS

TOPIC 1 STOICHIOMETRIC RELATIONSHIPS. 1.3 REACTING MASSES AND VOLUMES. ESSENTIAL IDEA. Mole ratios in chemical equations can be used to calculate reacting ratios by mass and gas volume. NATURE OF SCIENCE (1.8)

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TOPIC 1 STOICHIOMETRIC RELATIONSHIPS

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  1. TOPIC 1 STOICHIOMETRIC RELATIONSHIPS 1.3 REACTING MASSES AND VOLUMES

  2. ESSENTIAL IDEA Mole ratios in chemical equations can be used to calculate reacting ratios by mass and gas volume. NATURE OF SCIENCE (1.8) Making careful observations and obtaining evidence for scientific theories – Avogadro’s initial hypothesis.

  3. STOICHIOMETRY • For any stoichiometry problem, you must know what unit you are given and what unit you are looking for. • Step one is to ALWAYS convert to moles unless you are already in moles. • Step two is to multiply by the mole ratio with the unknown on top. • Step three is to convert to the unit you are looking for using the mole conversion chart.

  4. Stoich Problems – Mole to Mole • Given moles – looking for moles 2H2 + O2 2H2O How many moles of H2O are produced from 6.0 moles of oxygen? Step 1: Label your equation Step 2: Write down the given and multiply by the mole ratio with the unknown on top.

  5. 1. N2 + 3H2 2 NH3 How many moles of N2 are needed to make 12.2 moles of NH3? 2. 2C2H2 + 5O2 4CO2 + 2H2O How many moles of CO2 are produced from .80 moles of O2? 3. 2H2S+ 3O2 2SO2 + 2H2O How many moles of H2S must react with .68 moles O2? MOLE TO MOLE EXAMPLES

  6. PRACTICE - MOLE TO MOLE • 2C2H2 + 5O2 4CO2 + 2H2O • 1. How many moles of CO2 are produced when 14.3 moles of O2 are burned? • 2. How many moles of water are produced when .8432 moles of CO2 are produced? • 3. How many moles of oxygen are needed to react with 11.44 moles of C2H2?

  7. Stoich Problems – Mole to Mass • Given moles – looking for grams 2H2 + O2 2H2O How many grams of water are formed from 6.0 moles of H2? Step 1: Label the equation correctly. Step 2: Write down the given then multiply by the mole ratio – unknown on top. Step 3: Multiply by molar mass of unknown to convert from moles to grams.

  8. MOLE TO MASS EXAMPLES 2H2 + O2 2H2O • 1. How many grams of water are produced when 4.84 mol hydrogen reacts? • 2. How many grams of oxygen are need to react with 54.2 mol of hydrogen? • 3. How many grams of hydrogen are needed to produce 1.634 mol of water?

  9. PRACTICE – MOLE TO MASS • Ca3(PO4)2 + 3SiO2 + 5C 3CaSiO3 + 5CO + 2P • 1. How many grams of SiO2 are needed to produce 10.0 mol CO? • 2. How many grams of carbon are needed to react with 3.0 mol of Ca3(PO4)2? • 3. 42.0 mol of C produces how many grams of CaSiO3?

  10. Stoich Problems – Mass to Mole • Given grams – looking for moles 2H2 + O2 2H2O How many moles of water are formed from 16.0 grams of H2? Step 1: Label the equation correctly. Step 2: Put down the given and divide by molar mass of the given to get moles. Step 3: Multiply by mole ratio – unknown on top – to get moles.

  11. MASS TO MOLE EXAMPLES 2H2 + O2 2H2O • 1. How many moles of water are produced when 4.84 grams of hydrogen reacts? • 2. How many moles of oxygen are need to react with 54.2 grams of hydrogen? • 3. How many moles of hydrogen are needed to produce 1.634 grams of water?

  12. PRACTICE – MASS TO MOLE • Ca3(PO4)2 + 3SiO2 + 5C 3CaSiO3 + 5CO + 2P • 1. How many moles of SiO2 are needed to produce 420 g of CO? • 2. How many moles of CaSiO3 are produced when 100.0 g of Ca3(PO4)2 are reacted? • 3. How many moles of Ca3(PO4)2 are needed to produce 280.0 g of CO?

  13. Stoich Problems – Mass to Mass • Given grams – looking for grams 2H2 + O2 2H2O How many grams of O2 are needed to react with 20.0 g of H2 ? Step 1: Label the equation. Step 2: Write down the given and divide by the molar mass of the given. Step 3: Multiply by the mole ratio to find moles of unknown. Step 4: Multiply by molar mass of unknown to find grams of unknown.

  14. MASS TO MASS EXAMPLES 2H2 + O2 2H2O • 1. How many grams of water are produced when 4.84 grams of hydrogen reacts? • 2. How many grams of oxygen are need to react with 54.2 grams of hydrogen? • 3. How many grams of hydrogen are needed to produce 1.634 grams of water?

  15. PRACTICE – MASS TO MASS • N2 + 3H2 2NH3 • 1. How many grams of ammonia are formed from 15.0 g N2? • 2. How many grams of H2 are needed to react with 48.3 g N2? • 3. How many grams of H2 are needed to produce .914 g NH3?

  16. UNDERSTANDINGS/KEY IDEA1.3.A Reactants can be either limiting or excess.

  17. APPLICATION/SKILLS Be able to solve problems relating to reacting quantities, limiting and excess reactants, theoretical, experimental and percentage yields.

  18. Limiting Reactant • The reactant that will run out first in a reaction. • The excess reactant is the reactant that is not used up completely in reaction.

  19. Limiting Reactant • You will recognize a limiting reactant problem because there will be 2 “givens” in the problem. Example 2H2 + O2 2H2O If 4 moles of H2 react with 8 moles of O2 , how much water will be formed?

  20. Limiting Reactant - Calculations Step 1: Write down and convert both givens to moles (if not already in moles). Step 2: Set up two stoichiometric problems with the opposite reactant as the mole ratio. 1 mol O2 4 mols H2 given X = 2 mols O2needed 2 mol H2 2 mol H2 8 mols O2 given X = 16 mols H2needed 1 mol O2 Step 3: Interpret the equations by crossing and comparing to determine the limiting reactant. You need 2 moles of O2 and you have 8 mols O2 given You need 16 moles of H2, but you were only given 4 moles H2. Since you do not have enough H2 given, it will run out first. Therefore, H2 is the limiting reactant. Step 4: Use the limiting reactant, calculate the answer.

  21. 2H2(g) + O2(g)  2H2O(l) • How many grams of water are formed when 12.0 g of hydrogen reacts with 17.0 g of oxygen? • First of all, I recognize that I am given an amount of hydrogen and an amount of oxygen so I need to figure out which one is going to run out first and stop the reaction. This one is called the limiting reactant. The other reactant will be the excess reactant. • 12.0 g H2 x mol = 5.94 mol H2 given 2.02g • 17.0 g O2 x mol = .531 mol O2 given 32.0g • Notice that my first step was to convert to moles and to actually stop after the conversion to moles and write the word “given”.

  22. 2H2(g) + O2(g)  2H2O(l) • The next step is to find out how much of each reactant is needed to react with each other. You do this by multiplying each reactant by the mole ratio with each other. 12.0 g H2 x mol= 5.94 mol H2 given x 1 mol O2 = 2.97 mol O2 needed 2.02g 2 mol H2 17.0 g O2 x mol = .531 mol O2 given x 2 mol H2= 1.06 mol H2 needed 32.0g 1 mol O2 • Now cross compare to find the reactant that you do not have enough of. You need 1.06 mol H2 and you are given 5.94 mol H2 so you have more than enough. You need 2.97 mol O2 and are only given .531 mol O2 so you do not have nearly enough. This means that oxygen is your limiting reactant; therefore, hydrogen is your excess reactant.

  23. 2H2(g) + O2(g)  2H2O(l) • To solve the problem, you use the limiting reactant. (Important! Be sure to use the given moles, not the needed moles.) • The problem asked you for the mass of water produced. You now have a mole – mass problem. • .531 mol O2 x 2 mol H2O x 18.02g = 19.1 g H2O 1 mol O2 mol • The answer is your theoretical yield. • If you were asked how much excess reactant remained, simply subtract the moles H2 needed by the moles given. • 5.94 mol H2 given – 1.06 mol H2 needed = 4.88 mol H2 in excess.

  24. LIMITING REACTANT EXAMPLE • If you have 6.70 mol Na reacting with 3.20 mol Cl2, what is your limiting reactant, how many moles of product will be formed and how much excess reactant remains? • 2Na + Cl2 2NaCl

  25. UNDERSTANDINGS/KEY IDEA1.3.B The experimental yield can be different from the theoretical yield.

  26. Theoretical Yield – the maximum amount of product that can be produced from a given amount of reactant (found by using stoichiometry). Experimental Yield – the measured amount of product obtained from a reaction (what you got in the lab)

  27. Percent Yield Experimental Yield % Yield = X 100 Theoretical Yield

  28. You will recognize this type of problem when they ask you to find the percent yield. The experimental value will be given in the problem and you have to calculate the theoretical yield by using stoichiometry.

  29. EXAMPLE 2H2(g) + O2(g)  2H2O(l) • If 17.0 g of oxygen reacts to form 18.7 g of water, what is the percent yield? • First of all, you recognize that this is a percent yield problem. Circle 18.7 g and label it “experimental”. Next work the problem to solve for the theoretical yield. • 17.0 g O2 x mol x 2 mol H2O x 18.02 g = 19.1 g 32.0 g 1 mol O2 mol • % yield = exp x 100% = 18.7g x 100% = 97.9% theo 19.1g

  30. UNDERSTANDINGS/KEY IDEA1.3.C Avogadro’s law enables the mole ratio of reacting gases to be determined from volumes of the gases.

  31. Avogadro’s hypothesis states that equal volumes of different gases contain equal numbers of particles at the same temperature and pressure. This means that if you are given gas volumes and asked for gas volumes at the same conditions, you can find your answer using the mole ratios without doing any conversions.

  32. APPLICATION/SKILLS Be able to calculate reacting volumes of gases using Avogadro’s law.

  33. Example Problem 40 cm3 of carbon monoxide is reacted with 40 cm3 of oxygen in the following reaction. 2CO + O2 → 2CO2 What volume of carbon dioxide is produced? This is actually a limiting reactant “volumes of gases” problem and is quite often assessed on the IB exam.

  34. 40 cm3CO given x 1 mol O2 = 20 cm3O2needed 2 molCO 40 cm3O2given x 2 molCO= 80 cm3CO needed 1 mol O2 Notice you can use the mole ratio directly without converting cm3 to moles first. You need 80 cm3 CO and only have 40cm3 so CO is the limiting reactant. 40 cm3 CO given x 2 molCO2= 40 cm3CO2 produced 2 mol CO Oxygen is in excess by 20 cm3.

  35. UNDERSTANDINGS/KEY IDEA1.3.D The molar volume of an ideal gas is a constant at specified temperature and pressure.

  36. Standard Temperature and Pressure STP = 273 K (0°C) and 101.3 kPa (1atm) (Use 100 kPa when you can’t use a calculator.) Room Temperature and Pressure RTP = 298 K (25°C) and 101.3 kPa The molar volume of a gas at STP is 22.4 dm3/mol. The molar volume of a gas at RTP is 24 dm3/mol.

  37. APPLICATION/SKILLS Be able to solve problems and analyze graphs involving the relationship between temperature, pressure and volume for a fixed mass of an ideal gas.

  38. KINETIC THEORY OF MATTER • 1. The volume of the gas particles is assumed to be zero. • 2. The gas particles are in constant motion. • 3. The collisions of the gas particles with the sides of the container cause pressure. • 4. The particles exert no forces on each other. • 5. The average kinetic energy is directly proportional to the Kelvin temperature of the gas.

  39. PROPERTIES OF GASES • No definite shape • No definite volume • Very easily compressed • High rate of diffusion • Gas particles exert pressure on their surroundings.

  40. It is important to remember that you ALWAYS use Kelvin temperatures when working gas law problems. K = oC + 273 • Zero K (-273 oC) is called absolute zero and this is the temperature when all motion theoretically ceases to exist.

  41. BOYLE’S LAW • Robert Boyle, an Irish chemist, discovered that the volume of a gas was inversely proportional to the pressure applied. In other words, as pressure is increased, volume decreases. P1V1 = P2V2

  42. Graph of pressure vs volume volume pressure pressure 1/volume Boyle’s Law: The pressure of a gas is inversely proportional to the volume.

  43. CHARLES’S LAW • Jacques Charles, a French physicist, discovered the relationship between temperature and volume at constant pressure. He was the first person to make a solo balloon flight and it was on his flight that he discovered that the volume of a gas was dependent upon the temperature. • He discovered that the volume of a gas is directly proportional to the Kelvin temperature. In other words, if you increase the temperature, the volume increases. T1V2 = T2V1

  44. Graph of volume vs temperature volume Temperature (K) Charles’ Law – the volume of a gas is proportional to the Kelvin temperature.

  45. GAY LUSSAC’S LAW • Joseph Gay Lussac, a French physicist and chemist, another avid balloonist discovered the relationship between temperature and pressure. • He discovered that at constant volume, if you increase the temperature, the pressure increases. • He also invented many types of chemical glassware that is still in use today. T1P2 = P1T2

  46. Graph of pressure vs temperature pressure pressure Temperature (°C) Temperature (K) Gay Lussac’s Law – the pressure of a gas is directly proportional to the Kelvin temperature.

  47. Effect of temperature on gas volume • 1. If you double the temperature and keep volume constant, the pressure doubles. • 2. If you then double the volume and keep the temperature constant, the pressure is halved. • 3. Therefore, the net overall effect of doubling the temperature and doubling the volume is that there is no overall change in pressure.

  48. AVOGADRO’S LAW • Amadeo Avogadro, an Italian chemist, postulated that equal volumes of gases at the same temperature and pressure contained the same number of particles. • So for a gas at constant temperature and pressure, the volume is directly proportional to the number of moles of a gas. V1n2 = n1V2

  49. COMBINED GAS LAW • You can combine the first three gas laws into one equation. Remember to always use Kelvin temperature. • In this equation, you will be given 5 knowns and solve for 1 unknown. P1V1T2 = P2V2T1

  50. APPLICATION/SKILLS Be able to solve problems relating to the ideal gas equation.

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