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Stoichiometry

Stoichiometry. Stoichiometry. Needs a balanced equation Use the balanced equation to predict ending and / or starting amounts Coefficients are now mole ratios. In terms of Moles. The coefficients tell us how many moles of each kind.

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Stoichiometry

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  1. Stoichiometry

  2. Stoichiometry • Needs a balanced equation • Use the balanced equation to predict ending and / or starting amounts • Coefficients are now mole ratios

  3. In terms of Moles • The coefficients tell us how many moles of each kind. • Mole ratio- conversion ratio that relates the amounts in moles of any two substances in a chemical reaction. • Molar mass - mass, in grams, of one mole of a substance.

  4. Review: Molar Mass A substance’s molar mass (molecular weight) is the mass in grams of one mole of the compound. CO2 = 44.01 grams per mole H2O = 18.02 grams per mole Ca(OH)2 = 74.10 grams per mole

  5. 2 mol Al2O3 4 mol Al 2 mol Al2O3 3 mol O2 4 mol Al 3 mol O2 Mole Ratios 2 Al2O3(l) 4 Al(s) + 3 O2(g) Mole Ratios Mole Ratio (Fraction) 2 mol Al2O3 : 4 mol Al 2 mol Al2O3 : 3 mol O2 4 mol Al : 3 mol O2

  6. aA + bB cC + dD Atoms Atoms A B Molecules Molecules Formula Units Formula Units III. Stoichiometric “road map” (Use the balanced chemical equation) Mol Ratio Using the Coefficients from the balanced chemical equation Mass B Mass A Molar mass Molar mass Mol A Mol B 6.022 x 1023 6.022 x 1023

  7. 3 A + 4 B  2 D + 1 F How many moles of F are produced from 1.00 mol of A? 1 mol A 1 mol F = 0.33 mol F 3 mol A How many moles of D are produced from 5.00 mol of B? Mass A Mass B Molar mass Molar mass Mol Ratio 5 mol B 2 mol D MolA MolB 6.022 x 1023 6.022 x 1023 4 mol B Atoms A Atoms B Molecules Molecules Molecules = 2.50 mol D

  8. Mass A Mass B Molar mass Molar mass Mol Ratio MolA MolB 6.022 x 1023 6.022 x 1023 Atoms A Atoms B Molecules Molecules How many moles of lithium carbonate are produced when 5.3 mol CO2 are reacted? CO2(g) + 2LiOH(s)  Li2CO3(s) + H2O(l) 1. What is your starting point? 5.3 mol of CO2 2. What is your ending point? mol of Li2CO3 1 mol Li2CO3 5.3 mol CO2 1 mol CO2 = 5.3 mol Li2CO3

  9. Mass A Mass B Molar mass Molar mass Mol Ratio MolA MolB 6.022 x 1023 6.022 x 1023 Atoms A Atoms B Molecules 3 A + 4 B  2 D + 1 F How many grams of F are produced from 1.00 mol of A? If MM of F is 10.0g/mol. 10 g F 1 mol A 1 mol F = 3.33g F 3 mol A 1 mol F How many grams of D are produced from 5.00 mol of B? MM of D is 20.0g/mol 5 mol B 2 mol D 20 g D 4 mol B 1 mol D Molecules = 50.0g D

  10. Mass A Mass B Molar mass Molar mass Mol Ratio MolA MolB 6.022 x 1023 6.022 x 1023 A Atoms Atoms B Molecules Molecules What is the mass of glucose (C6H12O6) produced from 3.00 mol of water (H2O)? 6CO2(g) + 6H2O(l)  C6H12O6(s) + 6O2(g) 1. What is your starting point? 3.00 mol of H2O 2. What is your ending point? g of C6H12O6 180.81g C6H12O6 1 mol C6H12O6 3 mol H2O 6 mol H2O 1 mol C6H12O6 =90.1 g C6H12O6

  11. Mass A Mass B Molar mass Molar mass Mol Ratio MolA MolB 6.022 x 1023 6.022 x 1023 Atoms A Atoms B Molecules Molecules 6CO2(g) + 6H2O(l)  C6H12O6(s) + 6O2(g) What is the mass of oxygen (O2) produced from 2.50 mol of water (H2O)? 1. What is your starting point? 2.50 mol of H2O 2. What is your ending point? g of O2 6 mol O2 32.0 g O2 2.5 mol H2O 6 mol H2O 1 mol O2 =80.0 g O2

  12. Mass A Mass B Molar mass Molar mass Mol Ratio MolA MolB 6.022 x 1023 6.022 x 1023 Atoms A Atoms B Molecules Molecules 4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g) How many moles of NO are formed from 824 g of NH3? 1. What is your starting point? 824 g of NH3 2. What is your ending point? mol of NO 1 mol NH3 824 g NH3 4 mol NO 17.03g 4 mol NH3 = 48.4 mol NO

  13. Mass A Mass B Molar mass Molar mass Mol Ratio MolA MolB 6.022 x 1023 Atoms A Atoms B Molecules 3 A + 4 B  2 D + 1 F How many grams of F are produced from 5.00g of A? If MM of F is 10.0g/mol and MM of A is 25.0g/mol 5 g A 1 mole A 1 mole F 10 g F = 0.67g F 25 g A 3 mole A 1 mole F How many grams of D are produced from 5.00g of B? MM of D is 20.0g and MM of B is 10.0g/mol 6.022 x 1023 5 g B 20 g D 1 mole B 2 mole D Molecules 10 g B 4 mole B 1 mole D =5.00g D

  14. Mass A Mass B Molar mass Molar mass Mol Ratio MolA MolB 6.022 x 1023 6.022 x 1023 A Atoms Atoms B Molecules Molecules Sn(s) + 2HF(g)  SnF2(s) + H2(g) How many grams of SnF2 are produced from the reaction of 30.00 g HF? 1. What is your starting point? 30.00 g of HF 2. What is your ending point? g SnF2 30.00g HF 1 mol HF 1 mol SnF2 156.71 g SnF2 20.01g HF 2 mol HF 1 mol SnF2 = 117.5g SnF2

  15. Working a Stoichiometry Problem 6.50 grams of aluminum reacts with an excess of oxygen. How many grams of aluminum oxide are formed? 1. Identify reactants and products and write the balanced equation. 4 Al + 3 O2 2 Al2O3 a. Every reaction needs a yield sign! b. What are the reactants? c. What are the products? d. What are the balanced coefficients?

  16. Mass A Mass B Molar mass Molar mass Mol Ratio MolA MolB 6.022 x 1023 6.022 x 1023 Atoms A Atoms B Molecules Molecules 4 Al + 3 O2 2 Al2O3 6.50 g of Aluminum 1. What is your starting point? 2. What is your ending point? g of aluminum oxide 6.50 g Al 1 mol Al 2 mol Al2O3 101.96 g Al2O3 = ? g Al2O3 4 mol Al 1 mol Al2O3 26.98 g Al 6.50 x 2 x 101.96÷ 26.98 ÷ 4 = 12.3 g Al2O3

  17. Mass A Mass B Molar mass Molar mass Mol Ratio MolA MolB 6.022 x 1023 6.022 x 1023 Atoms A Atoms B Molecules Molecules Sn(s) + 2HF(g)  SnF2(s) + H2(g) How many grams of HF are produced from the reaction of 150.5 g H2? 150.5 g H2 1 mol H2 2 mol HF 20.01g HF 2.02g H2 1 mol H2 1 mol HF = 2982 g HF

  18. balanced II. Limiting Reagent A. Stoichiometric amounts: The proportions indicated in the _________ rxn. B. Most reactions do not have stoichiometric amounts. Generally, one reactant will be _________ before the other. The reactant that is depleted first is known as the ___________________. The reactant that is left at the end of the reaction is called the ___________________. C. Analogy: How to make a cheese sandwich. 2 slices of bread + 1 slice of cheese → 1 cheese sandwich   If you have 8 slices of bread and 6 slices of cheese, how many sandwiches can you make? __ (theoretical yield)   What is the limiting reagent? ______   What is the excess reagent? _______   How much of the excess reagent is left at the end of the rxn? ______________ depleted limiting reagent (LR) excess reagent (ER) 4 bread cheese 2 slices cheese

  19. limiting reagent D. Theoretical yield: The amount of product in grams that forms if all of the ________________ has reacted. (This number is CALCULATED on paper! Units are in grams only!) E. Actual yield: The amount of product (in grams) that is actually made (This number is from the EXPERIMENT). F. Percent yield: The comparison of the actual yield to the theoretical yield.

  20. Zn + 2 HCl  ZnCl2 + H2 If you have 1 mol of Zn, how much H2 would you make? 1 mol Zn 1 mol H2 = 1 mol H2 1 mol Zn If you have 1 mol of HCl, how much H2 would you make? 1 mol HCl 1 mol H2 = 0.5 mol H2 2 mol HCl What is the limiting reagent? HCl How much H2 is produced? 0.5 mol – theoretical yield

  21. Zn + 2 HCl  ZnCl2 + H2 If you have 0.25 mol of Zn, how much H2 would you make? 0.25 mol Zn 1 mol H2 = 0.25 mol H2 1 mol Zn If you have 1.00 mol of HCl, how much H2 would you make? 1.00 mol HCl 1 mol H2 = 0.5 mol H2 2 mol HCl Zn What is the limiting reagent? How much H2 is produced? 0.25 mol - theoretical yield

  22. PCl3 + 3 H2O  H3PO3 + 3 HCl 3.00 mol PCl3 and 3.00 mol H2O react. Determine the limiting reactant and theoretical yield of HCl. 1. Determine the limiting reactant 3.00 mol PCl3 3 mol HCl = 9.00 mol HCl  EXCESS 1 mol PCl3 3.00 mol H2O 3 mol HCl = 3.00 mol HCl  LIMITING 3 mol H2O 2. Determine the theoretical yield of HCl 3.00 mol

  23. PCl3 + 3 H2O  H3PO3 + 3 HCl 2. Determine the theoretical yield of HCl 3.00 mol 3. Determine the theoretical yield of HCl in grams 3.00 mol HCl 36.46g HCl = 109 g HCl 1 mol HCl

  24. PCl3 + 3 H2O  H3PO3 + 3 HCl Determine the limiting reactant and theoretical yield (g) of H3PO3 if 225 g of PCl3 and 123 g of H2O are reacted. 1. Determine the limiting reactant LIMITING 225 g PCl3 1 mol PCl3 82.00g H3PO3 1 mol H3PO3 = 134g H3PO3 137.32g PCl3 1 mol PCl3 1 mol H3PO3 EXCESS 123 g H2O 1 mol H2O 82.00g H3PO3 1 mol H3PO3 = 187g H3PO3 18.02g H2O 3 mol H2O 1 mol H3PO3

  25. PCl3 + 3 H2O  H3PO3 + 3 HCl The theoretical yield of this reaction is 134g H3PO3. However, the actual yield from the experiment is 120g. Calculate the percent yield. 120 g H3PO4 X 100% = 89.6 % 134 g H3PO4

  26. N2 + 3 H2 2 NH3 Determine the limiting reagent, the theoretical yield and the percentage yield if 14.0g N2 are mixed with 9.0g H2 and the 16.1g NH3 actually formed. LIMITING 14 g N2 1 mol N2 17.04g NH3 2 mol NH3 = 17.0g NH3 1 mol NH3 28.01g N2 1 mol N2 EXCESS 9 g H2 1 mol H2 17.04g NH3 2 mol NH3 = 50.6g NH3 1 mol NH3 2.02g H2 3 mol H2 16.1 g NH3 X 100% = 94.2 % 17.0 g NH3

  27. 16.1g of bromine are mixed with 8.42g of chlorine to give an actual yield of 21.1g of bromine monochloride. Determine the limiting reactant and the percentage yield. Br2 + Cl2  2 BrCl LIMITING 16.1 g Br2 1 mol Br2 115.35g BrCl 2 mol BrCl = 23.2g BrCl 1 mol BrCl 159.8g Br2 1 mol Br2 EXCESS 8.42 g Cl2 1 mol Cl2 115.35g BrCl 2 mol BrCl = 27.4g BrCl 1 mol BrCl 70.9g Cl2 1 mol Cl2 21.1 g BrCl X 100% = 90.9 % 23.2 g BrCl

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