1 / 32

O.A # 11

O.A # 11. Mole Ratio. Objective. Determine the mole ratios from a balanced equation. Page # 7 Mole to Mole Stoichiometry Mole Ratio is a ratio between the number of moles of any 2 substances in a chemical equation. Stoichiometric Calculations.

branxton
Télécharger la présentation

O.A # 11

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. O.A # 11 Mole Ratio

  2. Objective Determine the mole ratios from a balanced equation

  3. Page # 7 Mole to Mole Stoichiometry Mole Ratio is a ratio between the number of moles of any 2 substances in a chemical equation Stoichiometric Calculations All calculations begin with a balanced chemical equation which indicates the relative amounts of reactants and products

  4. Conversion factor to calculate moles of unknown moles given xcoefficient of unknown coefficient of given 1) How many moles of N2O and H2O are produced from 2.25 moles of NH4NO3? 2 NH4NO3→ N2O + H2O Moles______ ______ ______ 2.25 2.25 moles NH4NO3 x 1 N2O 1 NH4NO3 = 2.25 moles N2O 2.25 moles NH4NO3 x 2H2O 1 NH4NO3 = 4.50 moles H2O

  5. 2) How many moles of K and H2O are needed to produced 3.5 moles of H2 ? 2 2 2 K + H2O KOH + H2 3.5 Moles______ ______ ____________ 3.5 moles H2 x 2 K = 1 H2 7.0 moles K 3.5 moles H2 x 2 H2O = 1 H2 7.0 moles H2O

  6. Page 359 # 9-10 9) SO2 + H2O + O2H2SO4 2 2 2 12.5 Moles______ ______ ____________ 6.25 O2 12.5 H2SO4 10) 2CH4 + S82CS2 + 4H2S 1.50 Moles______ ______ ____________ 3.00 CS2 6.00 H2S

  7. O.A # 12 Find moles of Na needed to produce 12.6 moles of NaCl Na + Cl2 →NaCl 2 2 Moles ______ ______ ______ 12.6 12.6 12.6 moles NaCl x 2 Na 2 NaCl = 12.6 moles Na

  8. Objective Use the steps to solve stoichiometric problems

  9. Page # 8 Mole-to- Mass/Mass-to-Mass Stoichiometry 1. What mass of water is produced from 1.5 grams of glucose (C6H12O6)? C6H12O6 + 6O2 → 6CO2 + 6H2O Moles ______ ______ ____________ grams ______ ______ ____________ 0.0083 1.5 g Step 1: Balance chemical equation Step 2: Two sets of lines, Fill in given Step 3:Get to top line(change grams to moles) 1.5 gx 1 Mole = 0.0083 moles 180.0 g

  10. C6H12O6 + 6O2 → 6CO2 + 6H2O Moles ______ ______ ____________ 0.050 0.0083 0.90 1.5 g grams ______ ______ ______ ______ Step 4: advance across using mole ratios (COEFFICIENTS) = 0.050 moles H2O 0.0083C6H12O6x 6 H2O 1 C6H12O6 Step 5:Change moles to grams 0.050 moles x 18.0 g = 1 mole 0.90 g

  11. 2) What mass (in grams) of copper is produced when 5.66 g of iron reacts with excess CuSO4 Fe + CuSO4 FeSO4 + Cu 0.101 Moles ______ ______ ______ ________ 0.101 COEFFICIENTS 6.42 5.66 grams ______ ______ ______ _____ 5.66 gramsFex 1mole 55.8 grams = 0.101 moles Fe 0.101 moles Fex 1 Cu 1 Fe = 0.101 moles Cu 0.101 moles Cu x 63.6 grams 1 mole = 6.42 grams Cu

  12. Page 360 # 11 -12 11. 177 g Cl2 12. 88.6 g Cl2

  13. O.A # 13 Stoichiometry

  14. O.A # 14 Find grams of Oxygen if you have 14.6 g of Na 4 ___Na + ___O2 2 ____Na2O COEFFICIENTS 0.159 Moles ______ ______ ______ 0.635 grams ______ ______ ______ 5.09 14.6 14.6 g x 1 Mole__ 23.0 g = 0.635 moles 0.635 moles Na x 1 O2 4 Na 0.159 moles x 32.0 g__ 1 mole = 5.09 g

  15. Quick Write Explain what information you must have in order to calculate the mass of product formed in a chemical reaction when given the grams of reactant (mass to mass stoichiometry) (Give details…..)

  16. O.A # 15 Limiting Reactant

  17. Objective Identify the limiting reactant and calculate the amount remaining after the reaction is complete.

  18. Page # 9 12.3 Limiting Reactants Usually in chemical rxns, one reactant is used up before the other and the rxn stops. 1 bike = 1 frame + 2 tires 3 4 7 Excess reactant Limiting reactant

  19. O2 + 2 Mg 2 MgO 4 moles 2 moles 6 moles Limiting reactant: Reactant that is used up first. Determines the amount of product Products are predicted by L.R always Excess Reactant: left-over reactant

  20. 1. Identify the limiting reactant when 5.0 grams of water reacts with 2.25 grams of sodium to produce sodium hydroxide and hydrogen gas. Step 1: Set up problem as before (stoichiometry) 2H2O + 2Na  2 NaOH + H2 Moles _____ ______ ____________ 0.28 0.0978 grams_____ ______ ____________ 5.0 2.25 Step 2:convert grams of each reactant to moles H2O: 5.0 g x 1 Mole 18.0 g Na: 2.25 g x 1 Mole = 23.0 g = 0.28 moles = 0.0978moles

  21. Step 3: divide moles of reactant by coefficient in balanced equation. Smallest value: L.R 2H2O + 2Na  2 NaOH + H2 COEFFICIENTS Moles _____ ______ ____________ 0.28 0.0978 0.0978 3.91 grams_____ ______ ____________ 5.0 2.25 E.R = H2O 0.28 = 0.14 2 0.0978 = 0.0489 2 L.R = Na How many grams of NaOH are produced? = 3.91 g NaOH 0.0978 moles NaOH x 40.0 g 1 mole

  22. 2. If 79.1 g of Zinc reacts with 96.7 g of HCl to produce ZnCl2 and H2. Find L.R, How much ZnCl2 is produced? Zn + 2HCl  ZnCl2 + H2 Moles _____ ______ ____________ 1.21 1.21 2.65 COEFFICIENTS grams_____ ______ ____________ 79.1 g 96.7 g Zn: 79.1 g x 1 Mole = 65.4 g 1.21 moles = 1.21 L.R : Zn 1 HCl: 96.7 g x 1 Mole = 36.5 g 2.65 moles = 1.33 2 1.21 moles x 136.4 g 1 mole = 165 g ZnCl2

  23. O.A # 16 Excess Reactant

  24. O.A # 17 Theoretical Yield

  25. O.A # 18 Actual Yield

  26. Objective Calculate the theoretical yield and percent yield for a chemical reaction

  27. Page # 10 12.4 Percent Yield Actual Yield:amount of product actually produced under “NORMAL” lab conditions Theoretical Yield:maximum amount of product that can be produced under “PERFECT” lab conditions (calculated from stoichiometric calculations) Percent Yield Is the ratio of actual yield to theoretical yield Percent Yield = Actual Yield (given)x 100 Theoretical Yield (stoichiometry)

  28. Determine the percent yield for the reaction between 2.80 grams of aluminum nitrate and excess sodium hydroxide if 0.966 grams of aluminum hydroxide is recovered. Actual Yield Al(NO3)3 + 3 NaOH  Al(OH)3 + 3NaNO3 Moles ______ ______ ______ _________ 0.0131 2.80 grams ______ ______ ______ ______ Step 1: Set up problem as before (stoichiometry) Step 2: Convert grams to moles 2.80 g x 1 Mole__ 213.0 g

  29. Al(NO3)3 + 3 NaOH  Al(OH)3 + 3NaNO3 Moles ______ ______ ______ _________ 0.0131 0.0131 COEFFICIENTS 2.80 grams ______ ______ ______ ______ 1.02 Step 3: advance across using mole ratios and change to grams to calculate theoretical yield (T.Y) 0.0131 moles x 78.0 g__ 1 mole = 1.02 g T.Y Step 4: Calculate % Yield = 0.966x 100 1.02 % Yield = Actual Yield (given) x 100 Theoretical Yield % Yield= 94.7%

  30. O.A # 19 Percent Yield

  31. Objective Measure the number of moles of iron consumed and copper produced in the reaction of iron with copper II chloride

  32. O.A # 20 If 26.5 g of B react with 55.6 g of F2 to produce BF3. Find: a) L.R b) mass of BF3 2 3 ___B + ___F2 ____BF3 2 Moles ________ ________ _________ 2.45 1.46 0.973 55.6 66.0 grams ________ ________ _________ 26.5 2.45 moles B: 26.5 g x 1 mole = 10.8 g = 1.22 2 1.46 moles F2: 55.6 g x 1 mole = 38.0 g = 0.488 L.R : F2 3 x 67.8 g = 66.0 g 1 mole 1.46 moles F2 x 2 BF3 = 0.973 moles 3 F2

More Related