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Chemical Quantities

Chemical Quantities. Chapter 9. 9.2 Mole-Mole Relationships. 2H 2 O (l) → 2H 2(g) + O 2(g) 2 moles of water breaks down into 2 moles of Hydrogen and 1 mole of Oxygen How many moles of products are formed when we break down 4 moles of water? Multiply everything by 2

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Chemical Quantities

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  1. Chemical Quantities Chapter 9

  2. 9.2 Mole-Mole Relationships • 2H2O(l)→ 2H2(g) + O2(g) • 2 moles of water breaks down into 2 moles of Hydrogen and 1 mole of Oxygen • How many moles of products are formed when we break down 4 moles of water? • Multiply everything by 2 • 2[2H2O(l)→ 2H2(g) + O2(g)] • 4H2O(l)→ 4H2(g) + 2O2(g)

  3. What about a number like 5.8 moles? • Make it easier by dividing everything by 2 so we can start with 1 mole of water • [2H2O(l)→ 2H2(g) + O2(g)]/2 • H2O(l)→ H2(g) + ½ O2(g) • Now simply multiply everything by 5.8 • 5.8[H2O(l)→ H2(g) + ½ O2(g)] • 5.8H2O(l)→ 5.8H2(g) + 2.9O2(g)

  4. What if we want to find moles required when there are 2 reactants? • C3H8(g) + 5O2(g)→ 3CO2(g) + 4H2O(g) • Calculate the number of moles of oxygen required to react exactly with 4.3 moles of propane, C3H8, in the above reaction • 4.3 moles of C3H8 requires how many moles of O2 • There is a 1:5 ratio • So 4.3(1) : 4.3(5) • 4.3 = 21.5 • 4.3 mol C3H8(g) + 21.5 O2(g) • How many moles of products are formed?

  5. 9.3 Mass Calculations • We can do the same sort of thing with the mass of the elements and compounds • Let’s use same chemical equation • C3H8(g) + 5O2(g)→ 3CO2(g) + 4H2O(g) • Instead of moles we will use mass which is more realistic • If we have 44.1 g of propane, how much oxygen will it react with?

  6. Let’s use what we know again • We can convert grams of propane to moles of propane • 1 mole of propane is 44.09 g of propane • So 44.1 g x 1 mole / 44.09 g = 1 mole • Know we need 5 moles of oxygen for every mole of propane • 1 mol C3H8 x 5 mol O2 / 1 mol C3H8 = 5 mol O2 • Now how many grams of O2 are in 5 mol • 5 mol x 32 g / 1 mol = 160 g

  7. 9.5 Mass Calculations: Comparing Two Reactions • NaHCO3(s) + HCl(aq)→ NaCl(aq) + H2O(l) + CO2(g) • Reaction of baking soda and stomach acid • Mg(OH)2(s) + 2HCl(aq) → 2H2O(l) + MgCl2(aq) • Reaction of Milk of Magnesia and stomach acid • Looking at these 2 reactions, which antacid can consume the most stomach acid, 1.0 g of NaHCO3 or 1.0 g of Mg(OH)2(s)?

  8. What are we trying to figure out? • Want to know which reacts with the most moles of HCl • So need to do a mole to mole ratio • First must change grams of each antacid to moles

  9. Find the number of moles is in each mass • The molar mass of NaHCO3 is 84.01g so 1.00 g NaHCO3 x 1 mole / 84.01 g = 0.0119 mol • The molar mass of Mg(OH)2 is 58.33 g so 1.00 g Mg(OH)2 x 1 mole / 58.33 g = 0.0171 mol • Now we use the mole ratio for each • 1 mole NaHCO3 / 1 mole HCl = 0.0119 mole / x • 1 mole Mg(OH)2 / 2 mole HCl = 0.0171 mole / x • When we do the math we see that • 0.0119 mole NaHCO3 reacts with 0.0119 mole HCl • 0.0171 mole Mg(OH)2 reacts with 0.0342 mole HCl

  10. So which 1 gram sample of antacid consumes the most stomach acid? • A 1 g sample of Milk of Magnesia will consume more acid

  11. 9.6 The Concept of Limiting Reactions • When you have multiple ingredients involved in producing something, one will run out first and cause a limit on how much product you can make • Remember the Sandwich making example!!

  12. CH4(g) + H2O(g)→ 3H2(g) + CO(g) • What mass of H2O(g) is needed for 249 g of CH4(g)? (same steps we’ve been using) • Need to make sure equation is balanced • Need to convert grams CH4(g) to moles CH4(g) • Need to compare moles CH4(g) to moles H2O(g) • Need to change moles H2O(g) to grams H2O(g)

  13. Let’s do the Problem • Is Equation Balanced? Yeah!! • Convert 249 g CH4 to moles CH4 • 249g x 1mole/16g = 15.5 mol CH4 • Change moles CH4 to moles H2O • 15.5 mol CH4 x 1 mol H2O / 1 mol CH4 = 15.5 mol H2O • As proportion 1 mol CH4/15.5 mol CH4 = 1 mol H2O/ ? mol H2O • Change moles H2O to grams H2O • 15.5 mol H2O x 18 g / 1 mol = 279 g H2O • So in order for us to use up 249 g methane we need to have 279 g water

  14. But what if we didn’t have the right amount of each? • Well, 1 of them would be a limiting reactant • To solve we will use the same steps as before with a little twist

  15. Steps for Solving Limiting Reactant Problems • Write and Balance Equation • Convert Masses of Reactants to Moles • Use Mole Ratios to Determine what is Limiting • Use Amount of Limiting Reactant and Mole Ratios to Compute the Number of Moles of Product • Convert Moles to Grams (if needed)

  16. Problem • 25,000 g Nitrogen gas and 5,000 g Hydrogen gas react to form ammonia. How much ammonia is made when reaction stops? • Step 1 • N2 + 3H2→ 2NH3 • Step 2 • 25,000 g N2 x 1 mole N2 / 28 g N2 = 892 mol N2 • 5,000 g H2 x 1 mole H2/ 2.0 g H2 = 2,500 mol H2

  17. Step 3 (just pick one) • 892 mol N2 x 3 mol H2 / 1 mol N2 = 2,680 mol H2 • Or • 2,500 mol H2 x 1 mol N2 / 3 mol H2 = 833.33 mol N2 • This tells us we need 2,680 mol H2 and 833.33 mol N2 to use up all the reactants, do we have these amounts? • No, we do not have enough H2 (only have 2,500 mol and need 2,680 mol) so it will run out = limiting reactant

  18. Step 4 • 2,500 mol H2 x 2 mol NH3/3mol H2 = 1,650 mol NH3 • Step 5 • 1,650 mol NH3 x 17 g NH3/ 1 mol NH3= 28,100 g NH3 • Practice !!

  19. 9.8 Percent Yield • Products stop forming when 1 reactant runs out • Theoretical Yield = how much you would get in a perfect world • Actual Yield = what you actually get • Percent Yield = comparing these two • Actual / theoretical x 100 = % yield

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