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Chapter 11

Chapter 11. Stoichiometry. Stoichiometry. The study of quantitative relationships between the amounts of reactants used and amounts of products formed by a chemical reaction Law of Conservation of Mass. How Much?.

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Chapter 11

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  1. Chapter 11 Stoichiometry

  2. Stoichiometry • The study of quantitative relationships between the amounts of reactants used and amounts of products formed by a chemical reaction • Law of Conservation of Mass

  3. How Much? • If you were making cookies, you would need the right amount of each ingredient to get the two dozen cookies the recipe makes. • When performing a chemical reaction it is important to know how much of each reactant is needed to get the desired amount of product(s).

  4. Balanced Equations • The first step in determining anything in stoichiometry is balancing the chemical equation – Let’s Review! N2 + H2 → NH3 3 2

  5. Practice • CH4 + O2 → CO2 + H2O • NaOH + CaBr2 → Ca(OH)2 + NaBr • H2 + Cl2 → HCl • Li + H2O → LiOH + H2

  6. Types of Relationships • Particle to Particle • Mole to Mole • Mass to Mass C3H8(g) + 5O2(g)→ 3CO2(g)+ 4H2O(g) 44.10 g of propane gas reacts with 159.99 g of oxygen gas to form 132.03 g of carbon dioxide gas and 72.06 g of water vapor. One molecule of propane gas reacts with five molecules of oxygen gas to form three molecules of carbon dioxide gas and four molecules of water vapor. One mole of propane gas reacts with five moles of oxygen gas to form three moles of carbon dioxide gas and four moles of water vapor.

  7. Practice 5. 2H2O(g) → 2H2(g)+ O2(g) 6. 2NaCl(aq) + MgBr2(aq) → MgCl2(s)+ 2NaBr(aq) 7. CaO(s) + CO2(g) → CaCO3(s)

  8. Mole Ratio • The ratio between the numbers of moles of any two substances in a balanced chemical equation. Taken from the coefficients of the balanced equation. 2K(s) + Br2(l) → 2KBr(s) 2 mol K 2 mol K 1 mol Br2 2 mol KBr 2 mol KBr 2 mol KBr 1 mol Br2 2 mol Br2 1 mol Br2 1 mol Br2 2 mol K 2 mol KBr

  9. Mole Ratio • For reactions with more reactants and products there are more mole ratios BaS(aq) + 2KOH(aq) → Ba(OH)2 + K2S 1 mol BaS 1 mol BaS 1 mol BaS 2 mol KOH 1 mol Ba(OH)2 1 mol K2S 2 mol KOH 2 mol KOH 2 mol KOH 1 mol BaS 1 mol Ba(OH)2 1 mol K2S

  10. Practice 8. H2 + Cl2 → HCl 9. H2O → H2 + O2 10. C2H10 + O2 → CO2 + H2O 11. LiOH + CaBR → Ca(OH)2 + LiBr

  11. Mole to Mole Conversions N2 + 3H2 → 2NH3 • If we have one mole of nitrogen gas, we need three moles of hydrogen gas in order to produce two moles of ammonia. • What if we have 0.012 mol of nitrogen gas?

  12. Mole to Mole Conversions N2 + 3H2 → 2NH3 • What if we have 0.012 mol of nitrogen gas? • Use mole ratios 3 mol H2 1 mol N2 2 mol NH3 1 mol N2 0.012 mol N2 × = 0.036 mol H2 = 0.024 mol H2 0.012 mol N2 ×

  13. Practice 12. 0.6 mol hydrogen gas H2 + Cl2 → HCl 13. 1.45 mol oxygen gas C2H10 + O2 → CO2 + H2O 14. 1.32 mol hydrogen gas H2O → H2 + O2 15. 0.24 mol lithium hydroxide LiOH + CaBr2→ Ca(OH)2 + LiBr

  14. Mole to Mass Conversions N2(g) + 3H2(g) → 2NH3(g) • What if we have 5.20 mol of nitrogen gas? How many grams of ammonia are produced in an excess of hydrogen gas? • Use mole ratios, then molar mass 2 mol NH3 1 mol N2 × 5.20 mol N2 = 15.6 mol NH3 17.03 g 1 mol NH3 × 15.6 mol NH3 = 266 g NH3

  15. Mole to Mass Conversions 4Al(s) + 3O2(g) → 2Al2O3(s) • What if we have 0.42 mol of oxygen gas? How many grams of aluminum oxide are produced in an excess of aluminum solid? • Use mole ratios, then molar mass 2 mol Al2O3 3 mol O2 × 0.42 mol O2 = 0.28mol Al2O3 101.96 g 1 mol Al2O3 × 0.28mol Al2O3 = 29 g Al2O3

  16. Practice 16. 0.6 mol chlorine gas Na(s) + Cl2(g) → NaCl(s) 17. 12 mol oxygen gas P(s) + O2(g) → P2O5(s) 18. 3.05 mol oxygen gas CH4(g) + O2(g) → CO2(g) + H2O(g) 19. 0.96 mol sodium bromide NaBr(s) + CaF2(s) → CaBr2(s) + NaF(s)

  17. Mass to Mass Conversions N2(g) + 3H2(g) → 2NH3(g) • What if we have 27 g of nitrogen gas? How many grams of ammonia are produced in an excess of hydrogen gas? • Use molar mass, then mole ratios, then molar mass 1 mol N2 28.01 g N2 × = 0.96 mol N2 27 g N2 2 mol NH3 1 mol N2 × = 1.92 mol NH3 0.96mol N2 17.03 g 1 mol NH3 × 1.92 mol NH3 = 33 g NH3

  18. Mass to Mass Conversions 4Al(s) + 3O2(g) → 2Al2O3(s) • What if we have 14.3 g of oxygen gas? How many grams of aluminum oxide are produced in an excess of aluminum solid? • Use molar mass, mole ratios, then molar mass 1 mol O2 32.00 g O2 × 14.3 g O2 = 0.447mol O2 2 mol Al2O3 3 mol O2 × 0.447 mol O2 = 0.298mol Al2O3 101.96 g 1 mol Al2O3 × 0.298mol Al2O3 = 30.4 g Al2O3

  19. Practice 20. 46 g chlorine gas Na(s) + Cl2(g) → NaCl(s) 21. 31.7 g oxygen gas P(s) + O2(g) → P2O5(s) 22. 3.05 g oxygen gas CH4(g) + O2(g) → CO2(g) + H2O(g) 23. 6.41 g sodium bromide NaBr(s) + CaF2(s) → CaBr2(s) + NaF(s)

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