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Stoichiometric Calculations

Stoichiometry –Ch. 9. Stoichiometric Calculations. A. Proportional Relationships. Ratio of eggs to cookies. 3/4 c. brown sugar 1 tsp vanilla extract 2 eggs 2 c. chocolate chips Makes 5 dozen cookies. 2 1/4 c. flour 1 tsp. baking soda 1 tsp. salt 1 c. butter 3/4 c. sugar.

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Stoichiometric Calculations

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  1. Stoichiometry –Ch. 9 Stoichiometric Calculations

  2. A. Proportional Relationships Ratio of eggs to cookies 3/4 c. brown sugar 1 tsp vanilla extract 2 eggs 2 c. chocolate chips Makes 5 dozen cookies. 2 1/4 c. flour 1 tsp. baking soda 1 tsp. salt 1 c. butter 3/4 c. sugar I have 5 eggs. How many cookies can I make? 5 eggs 5 doz. 2 eggs = 12.5 dozen cookies

  3. A. Definition Stoichiometry • Mass, volume, and particle number relationships between substances in a chemical reaction • based on the mole ratio Mole Ratio • indicated by coefficients in a balanced equation

  4. B. Uses of Stoichiometric Calculations Used to determine the mass, number of moles or particles, or volumes for any reactants or products in a reaction simply by knowing the quantity of one substance. This has a huge industrial significance.

  5. C. Coefficients are a Ratio _N2(g) + 3H2(g) 2NH3(g) 1 moleculeN2 + 3 moleculesH2  2 moleculesNH3 Ratio: 1, 3, 2 Multiply the ratio by 6.022 x 1023 (1 mole): 1 mole N2 + 3 moles H2  2 moles NH3

  6. D. Conversions to Remember Mole Relationships: 1 mole of anything contains 6.022 x 1023 of those things. 1 mole of any substance is equal to its molar mass in grams.

  7. D. Molar Relationships StandardTemperature&Pressure 0°C and 1 atm 1 mol of a gas=22.4 L at STP

  8. D. Molar Relationships LITERS OF GAS AT STP Molar Volume (22.4 L/mol) MASS IN GRAMS NUMBER OF PARTICLES MOLES Molar Mass (g/mol) 6.022  1023 particles/mol Molarity (mol/L) LITERS OF SOLUTION

  9. Mole-Mole Conversions Aluminum + oxygen  aluminum oxide 4Al(s) + 3O2(g)  2Al2O3 How many moles of Al are needed to form 2.3 moles of Al2O3? 2.3 mol Al2O34 molAl = 4.6 mol Al 2 molAl2O3

  10. Mole-Mole Conversions Aluminum + oxygen  aluminum oxide 4Al(s) + 3O2(g)  2Al2O3 How many moles of O2 are required to react completely with 0.84 moles Al ? 0.84 mol Al 3 mol O2 = 0.63 mol O2 4 molAl

  11. Mole-Mole Conversions Aluminum + oxygen  aluminum oxide 4Al(s) + 3O2(g)  2Al2O3 Calculate the number of moles of Al2O3 formed when 17.2 mol O2 reacts with Al. 17.2 mol O22 molAl2O3 = 11.5 mol 3 molO2 Al2O3

  12. General Procedure for solving: a. Write a balanced equation. b. Write the given in the first box. c. Line up conversion factors needed • Mole ratio - moles  moles • Molar mass - moles  grams • Molarity - moles  liters soln • Molar volume - moles  liters gas • Mole ratio - moles  moles Core step in all stoichiometry problems!! d. Everything should cancel out except the desired unit.

  13. Mass - Mass Problems Calculate the mass of HCl (aq) required to react with 10.0g of zinc metal. ? g 10.0 g 2HCl(aq) + Zn(s) ZnCl2(aq)+ H2(g)

  14. D. Mass – Mass Problems ? g 10.0 g 2HCl(aq) + Zn(s) ZnCl2(aq)+ H2(g) (mass moles) 1 mol Zn 65.39 g Zn (mass known) 10.0g Zn (mole ratio) 2 mol HCl 1 mol Zn (moles mass) 36.46 g HCl 1 mol HCl = 11.2 g HCl

  15. Mass - Mass Problems Calculate the mass of O2 gas produced if 2.50 g KClO3are completely decomposed by heating. ? g 2.50 g 2KClO3 2KCl + 3O2 2.50 g KClO3 1 mol KClO3 122.55 g KClO3 3 mol O2 2 mol KClO3 32.00 g O2 1 mol O2 = 0.979 g O2

  16. Mass - Gas Volume Problems What volume of hydrogen at STP can be produced from the reaction of 6.54g of zinc with hydrochloric acid? 6.54 g ? L 2HCl(aq) + Zn(s) ZnCl2(aq)+ H2(g) Instead of looking for a mass in grams, you are trying to find a volume in liters.

  17. Mass – Gas Volume Problems ? L 6.54 g 2HCl(aq) + Zn(s) ZnCl2(aq)+ H2(g) (mass moles) 1 mol Zn 65.39 g Zn (mass known) 6.54g Zn (mole ratio) 1mol H2 1 mol Zn (moles liters) 22.4 L H2 1 mol H2 = 2.24 L H2

  18. Mass – Gas Volume Problems Bromine will react with 5.60 x 103 ml of hydrogento yield what mass of hydrogen bromide at STP? ? g 5.60 x 103 ml = 40.5 g HBr Br2(g) + H2(g) 2HBr(g) 1 mol H2 22.4 L H2 2 mol HBr 1 mol H2 1 L H2 1000 mL H2 80.91 g HBr 1 mol HBr 5.60 x 103ml H2

  19. Volume – Volume Problems How many liters of oxygen are required to burn 1.00 L of methane gas, CH4, completely. ? L 1.00 L CH4(g) + 2O2(g) CO2(g) + 2H2O(g) 1.00 L CH4 1 mol CH4 22.4 L CH4 2 mol O2 1 mol CH4 22.4 L O2 1 mol O2 = 2.00 L O2

  20. Volume – Volume Problems What volume of bromine gas is produced if 75.2 L of chlorine gas react with excess HBr? ? L 75.2 L Cl2(g) + 2HBr(g) Br2(g) + 2HCl(g) 75.2 L Cl2 1 mol Cl2 22.4 L Cl2 1 mol Br2 1 mol Cl2 22.4 L Br2 1 mol Br2 = 75.2 L Br2

  21. Short Way (Vol-Vol only) • The ratio of combining volumes is the same as the ratio of the combining moles. • Therefore, use the coefficients as the ratios for volumes.

  22. Short Way (Vol-Vol only) • This can’t be applied to solids or liquids, because one mole of a solid or liquid may have a different volume than one mole of a different solid or liquid.

  23. Volume – Volume Problems How many liters of CO2 gas will be produced by burning completely 5.00 L of ethane, C2H6? 5.00L ? L 2C2H6(g) + 7O2(g) 4CO2(g) + 6H2O(g) 5.00 L C2H6 4 L CO2 2 L C2H6 = 10.0 L CO2

  24. Stoichiometry Problems How many grams of SnF2 can be made by reacting 7.42 x 1024 molecules of HF with tin according to the equation: 7.42 x 1024 molecules ? g Sn(s) + 2HF(g) SnF2(S)+ H2(g) 7.42 x 1024molecules HF 1 mol SnF2 2 mol HF 1 mol HF 6.022 x 1023 molecules HF 156.71 g SnF2 1 mol SnF2 = 965 g SnF2

  25. Stoichiometry Problems How many molecules of H2 are produced by the reaction of tin with 80.0 L of HF at STP according to the equation: 80.0 L ? molecules Sn(s) + 2HF(g) SnF2(S)+ H2(g) 6.022 x 1023 molecules H2 1 mol H2 1 mol H2 2 mol HF 1 mol HF 22.4 L HF 80.0 L HF = 1.08 x 1024 molecules H2

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