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STOICHIOMETRY

STOICHIOMETRY. Introduction. We will look at quantitative measurements in chemical reactions. Mass and mole relationships in chemical reactions Limiting reagents in a reaction Theoretical and percent yield Solution concentrations and preparations.

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STOICHIOMETRY

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  1. STOICHIOMETRY

  2. Introduction • We will look at quantitative measurements in chemical reactions. • Mass and mole relationships in chemical reactions • Limiting reagents in a reaction • Theoretical and percent yield • Solution concentrations and preparations

  3. Quantitative measurements are those that involve actual number relationships, not just human judgement. Ex. 3.21 cm, 5.42 moles, 342.65 g, 24.1 L not just, “It was really, really a lot”

  4. N2 + 3H2 -----> 2NH3 In a balanced chemical equation, the coefficients allow us to calculate how much of a substance is used, produced, or needed in the reaction. continued...

  5. N2(g)+ 3H2 (g)-----> 2NH3 • From this equation we can determine that 1 mole of Nitrogen gas (28.02 g ) reacts with 3 moles of Hydrogen gas (6.06 g) to produce 2 moles of ammonia (28.02g +6.06g=34.08g of ammonia)

  6. Ex. 1 How many moles of NaOH will react with 3.21 moles of sulfuric acid? Step 1. Always write the balanced chemical equation. Step 2. Write down the given numeric info. If it is not in mole units convert it to moles using the molecular mass as a conversion factor. continued....

  7. continued... Step 3. Now convert from the moles of starting substance to the moles of the desired substance by using a mole to mole ratio from the chemical reaction.

  8. continued... Step 4. If the problem is to find grams of the substance, convert the moles to grams using the molecular mass conversion factor. Step5. Check units and significant figures.

  9. continued... Now lets work out the example problem: Step 1. H2SO4 + 2NaOH -----> Na 2SO4 + 2HOH continued...

  10. continued... Step 3 (Note that step 2 is not needed) 3.21 moles H2SO4x 2moles NaOH 1 mol H2SO4 = 6.42 moles NaOH Are the significant figures correct?

  11. Ex. 2 How many moles of Hydrogen are required to react with 103.25 g of Nitrogen in the Haber Process to produce ammonia? Step 1. 3H2 + N2 ------> 2NH3 continued...

  12. continued... Step 2. 103.25 g N2 x 1mole N2 =3.68 moles 28.02 g N2 ofN2 continued...

  13. continued... Step 3. 3.68 moles N2 x 3 moles H2= 10.0mole 1 mol N2 H2 Are the significant figures correct?

  14. Ex. 3 How many grams of Hydrogen are required to react with 100.00g of Oxygen in the production of water? Step 1. 2H2 + O2 ----> 2H2O continued...

  15. continued... Step 2. 100.0 g O2 x 1 mole O2= 3.12mol O2 32.0 g O2 Step 3. 3.12 mol O2 x 2mol H2 = 6.24mol H2 1 mol O2.. continued...

  16. continued... Step 4. 6.24 mol H2 x 2.02 g H2=12.60 g H2 1 mol H2 Are the significant figures correct?

  17. Remember.... In multiplication and division, the number of significant figures in the answer should be equal to the least number of significant figures in the problem. Coefficients in the chemical equation are exact numbers.

  18. The Limiting Reagent We have already seen, from the chemical equations in the previous problems, that certainratios of substances will react. In a chemical reaction, one of the substances may be in excess of the needed ratio (or one of them may be short of the needed ratio)

  19. Notice that there is an excess of red atoms after the bonds have been made. There is a shortage of the green atoms. The green are the limiting reagent.

  20. Limiting Reagent continued... We must be able to calculate which substance is the limiting reagent because it will be the one to stop the reaction from continuing. Once we have done this, we can calculate, theoretically, the amount of product that should be produced.

  21. continued... For example, if you need4 slices of bread and 2 pieces of cheese to make 2 sandwiches and you have4 slices of bread and 1 piece of cheese, the cheese is the limiting reagent. Your product will be only 1 sandwich. The process of making sandwiches will stop before you get to make the second sandwich.

  22. A chemical reaction works in a similar fashion but we must calculate the quantities in question. Ex. 4 Iron combines with Sulfur according to the following reaction: Fe (s) + S(l) -----> FeS (s) In an experiment, 7.62 g Fe are allowed to react with 8.67 g S. continued....

  23. continued A. Determine which one of the reactants is the limiting reactant. b. Calculate the mass of FeS formed. A. 7.62g Fe x 1 mole Fe = 0.136 mol Fe 55.85 g Fe 8.67 g S x 1 mol S=0.270 mol S 32.06 g S continued...

  24. continued... We now know that we have 0.136 mol Fe and 0.270 mol S. Use the balanced chemical equation to determine how many moles of each will be needed to react with each other. continued...

  25. continued... 0.136 mol Fe x 1 mol S = 0.136 mol S 1 mol Fe 0.136 mol S needed to react with the Fe. 0.270 mol S x 1 mol Fe = 0.270 mol Fe 1 mol S 0.270 mol Fe needed to react with the S. continued...

  26. continued... Since we have 0.270 moles of S and only need 0.136 moles of it to completely react with the given amount of Fe, we clearly have an excess amount of S. Since we only have 0.136 moles of Fe and we need 0.270 moles of it to completely react with the given amount of S, the Fe is the reactant that will limit the reaction. It is the limiting reagent. continued...

  27. We have completed part A. Now that we know which substance will limit the reaction we can determine how much product, FeS, will be produced. B. Since 1 mole of FeS is produced from 1 mole of Fe, then 0.136moles Fe x 1 mol FeS = 0.136 mol 1 mol Fe FeS produced continued...

  28. Now calculate the mass of FeS... 0.136 mol FeS x 87.92 g FeS = 12.0 g 1 mol FeS FeS Since we had 0.270 moles S and 0.136 moles Fe, and the Fe was the limiting reagent, only 0.136 moles of the S could react to produce0.136 moles of FeS. That leaves 0.134 mole of S left unreacted (or 4.30 grams).

  29. Yields of Reactions In the previous example we calculated the amount of FeS that, theoretically would be produced from the reaction. In the laboratory, due to experimental error, the yield of product is usually less than that which is calculated on paper. The yield that is produced in the lab is called the “Actual Yield”.

  30. Percent Yield The percent yield is calculated by dividing the actual yield by the theoretical yield and then, because it is a percent, multiply by 100.

  31. Ex. 5 In the previous problem the theoretical yield of FeS was calculated to be 12.0 g. In a lab experiment the actual yield was 10.55 g. Calculate the % Yield. 10.55g x 100 = 87.92 % 12.0 g

  32. Solutions • Calculating molar concentration, (Molarity) • Preparing dilutions • Stoichiometry of Solutions

  33. Helpful formulas and definitions: M = Molarity = moles of solute liters of solution Solute is the substance in the lesser quantity which is dissolved in the substance of greater quantity. Solvent is the substance in greater quantity. Solutionis the combined solute plus solvent.

  34. To calculate a dilution... CcVc= CdVd C stands for concentration, Vstands for volume cstands for concentrate , d stands for dilute

  35. What this means is.... The concentration of the concentrated substance times the volume of the concentrated substance is equal to the concentration of the dilute substance times the volume of the dilute substance.

  36. Ex. 6 Dilution How would you prepare 5.00 x 10 2mL of 1.75 M HNO3 from 8.6 M stock solution? Cc Vc = Cd Vd (8.6 M)(V) = (1.75 M)(5.00x102 mL) Solving for V = 102 mL of 8.6 M substance

  37. So, to prepare the dilute solution... 102 mL of the more concentrated 8.6 M solution is diluted with water up to 5.00 x 10 2mL (500 mL) to produce the more dilute 1.75 M solution.

  38. Ex. 7 Molarity What is the molarity of a solution that has 3.21 moles of HCl dissolved in 2.4 L of solution? Molarity = moles of solute = 3.21 moles HCl Liters of solution 2.4 L solution = 1.34 M

  39. Ex. 8 Molarity How many grams of sodium sulfate (Na2SO4 ) are required to prepare a 250 mL, 0.683 M solution? Since molarity is in units of moles / liter we must change the mL into L : 250 mL x 1 L = 0.250 L 1000mL continued...

  40. continued... 0.250 L x 0.683 moles Na 2 SO 4 1 liter solution = 0.171 moles Na 2SO 4 0.171 moles Na2SO4 x 142.04 g Na 2SO4 1 mole Na2SO4 = 24.3 g Na2SO4

  41. Ex. 9 Stoichiometry In the reaction of Zn with HCl, given 10.0 g of Zn, what volume of 2.50 M HCl (in mL) would be needed to completely react? We follow the steps of solving a stoichiometry problem but we include the molarity ratio in order to get volume.

  42. continued... Step 1. Zn + 2HCl ----> ZnCl2 + H2 Step 2. 10.0 g Zn x 1 mol Zn = 0.153 mol Zn 65.39 g Zn continued....

  43. continued... Step 3. 0.157 mol Zn x 2 mol HCl =0.306mol HCl 1 mol Zn Step 4. 0.306 mol HCl x 1.00 L soln. = 0.122 L 2.50 mol HCl HCl soln. = 122 mL HCl soln.

  44. You should be able to... • Do Stoichiometry calculations involving : • 1. Mole - Mole relationships • 2. Mass - Mole relationships • 3. Mass - Mass relationships • 4 Solutions • Calculate the theoretical and % Yield • Determine Molarity of a solution and prepare it • Prepare a diluted solution • Determine the limiting reagent

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