1 / 22

Topic 8.3 Stoichiometry: Limiting and Excess Reagent Calculations

Topic 8.3 Stoichiometry: Limiting and Excess Reagent Calculations. Page 320-327. By Kirsten. What is Stoichiometry? Lets take a closer look. A method of predicting or analyzing the quantities of the reactants and products participating in a chemical process

zeroun
Télécharger la présentation

Topic 8.3 Stoichiometry: Limiting and Excess Reagent Calculations

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Topic 8.3 Stoichiometry: Limiting and Excess Reagent Calculations Page 320-327 By Kirsten

  2. What is Stoichiometry?Lets take a closer look. • A method of predicting or analyzing the quantities of the reactants and products participating in a chemical process • Three forms including gas stoichiometry, solution stoichiometry, and gravimetric stoichiometry.

  3. Example of Stoichiometry • In a precipitation reaction, KOH(aq) reacts with excess Sn(NO3)2(aq) to produce a precipitate. If the mass of precipitate is 2.57 g, what mass of KOH(s) was present in the original solution?

  4. First Step: Balance Equation: 2KOH(aq) + Sn(NO3) 2(aq) Sn(OH)2(s) + 2KNO3(aq)

  5. Second Step: n1= 2.57 g x (1 mol/ 152.71 g) = 0.0168 mol n2= 0.0168 mol x (2/1) = 0.0337 mol m= 0.0337 mol x (56.11 mol/g) = 1.89g

  6. Understanding Chemical Principles • Conservation of Mass in a Chemical Reaction • In chemical reactions the mass is conserved, meaning that the mass of the products equals the mass of the reactants. • Mass Products = Mass Reactants • There will always be an equal number of moles of each element before and after a reaction takes place

  7. Identifying Reagents • Limiting Reagents • Completely consumed in a chemical reaction • Excess Reagents • More is present than is necessary to react with the limiting reagent • The limited reagent is the reagent that is being analyzed in a quantitative analysis where limiting and excess reagents are present.

  8. How to find Limiting Regent: • Ensure that all masses given are in moles. • Take the molar amounts from each substance and divide by the coefficient of that substance in the balanced equation. • The smaller number will be the limiting reagent.

  9. Example 1: • 300 mL of 0.100 mol/L of BaCl2(aq) and 200 mL of 0.110 mol/L of Na2CO3(aq) are mixed. What is the limiting reagent in the reaction? Answer: Na2CO3

  10. Why? BaCl2(aq) + Na2CO3(aq) BaCO3(s) + 2NaCl(aq) With a 1:1 mole ratio of reactants he species present in least amount is the limiting reagent. nBaCl2= 300mL x (0.100mol/L) =30.0 mmol nNa2CO3= 200 mL x (0.110mol/L) = 22.0 mmol Therefore Na2CO3(aq) is the limiting reagent

  11. Example 2: • 100.0 g of iron (III) chloride and 50.00g of hydrogen sulfide react. What is the limiting reagent? 2FeCl3 + 3H2S Fe2S3 + 6HCl

  12. Answer: • Iron(III) chloride 100g/ 162.204 g/mol =0.6165 mol • Hydrogen Sulfide 50.00g/ 34.081 g/mol =1.467 mol 1.467 mol/ 3 =0.489 • Iron (III) chloride is the limiting reagent

  13. Example 3: • In an experiment, 26.8 g of iron(III) chloride in solution is combined with 21.5 g of sodium hydroxide in solution. Which reactant is in excess, and by how much? What mass of each product will be obtained?

  14. Answer: FeCl3(aq) + 3NaOH(aq) Fe(OH)3(s) + 3NaCl(aq) 26.8 g 21.5g m m 162.20g/mol 40.00 g/mol 106.88 g/mol 58.44 g/mol nFeCl3 = 26.8 g x (1mol/ 162.20g0 =0.165 mol nNaOH = 21.5 g x (1 mol / 40.00g) = 0.538 mol

  15. Answer: nNaOH=0.165 mol x (3/1) =0.496 nNaOH= 0.538 mol - 0.496 mol = 0.042 mol mNaOH = 0.042 mol x (40.00g/1 mol) = 1.7g

  16. Answer: mFe(OH)3= 0.165 molFeCl3 x (1 mol Fe(OH)3/1 molFeCl3) x (106.88 gFe(OH)3/ 1 mol Fe(OH)3) = 17.7 g Fe(OH)3 m NaCl = 0.165 molFeCl3 x (3 mol NaCl/ 1 mol FeCl3) x (58.44 g NaCl / 1 mol NaCl) = 29.0 NaCl Therefore sodium hydroxide is in excess by 1.7 g, the mass of iron (III) hydroxide produced is 17.7 g and the mass of sodium chloride produced is 29.0g.

  17. Theoretical yields Vs. Actual yields • The theoretical yield of a chemical reaction is the amount of product formed if all of the limiting reagent reacts. • Calculated using stoichiometry. • The actual yield is the actual quantity of products formed after a chemical reaction. • Usually the theoretical yield will be greater than the actual yield that is produced.

  18. Reasons for Discrepancy • The actual yield of a chemical reaction is usually less than the theoretical yield for these reasons • Purity of chemicals being used • Errors in measurements • Experimental factors that may have lead to loss of reactants

  19. % Error Calculations: • A reasonable quantity of reasonable excess reagent is 10%

  20. Example 4: • You decide to test the method of stoichiometry using the reaction of 2.00 g of copper(II) sulfate in solution with an excess of sodium hydroxide in solution. What would be a reasonable mass of sodium hydroxide to use? CuSO4(aq) + 2NaOH(aq) Cu(OH)2(aq) + Na2SO4(aq) 2.00 g m 159.62 g/mol 40.00 g/mol

  21. Answer: nCuSO4 = 2.00g x (1 mol/159.62 g) =0.0125 mol nNaOH =0.0125 mol x (2/1) =0.0251 mol mNaOH = 0.0251 mol x (40g/1mol) = 1.00 g Now add 10% to this amount to determine the reasonable mass of sodium hydroxide to use. 1.00 g + 0.10g =1.10g

  22. For More Information: • Nelson Chemistry Text • Pages 320-327 • The Key- Chemistry 20 • D2L Web Lessons

More Related