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Chapter 9 Chemical Quantities. In order to avoid waste and to be economic, a chemist has to know how much to make, a doctor must know how much to prescribe, an engineer has to know how much fuel to mix with others, etc., etc., etc...

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## Chapter 9 Chemical Quantities

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**In order to avoid waste and to be economic, a chemist has to**know how much to make, a doctor must know how much to prescribe, an engineer has to know how much fuel to mix with others, etc., etc., etc... • Questions we answer here in Ch 9? How much can I get?How much do I need? The engineers who develop airbags have to make sure exactly the right amount of gas is made - no more, no less.**Ch 9 is a compilation of Chs 1-8!**• must name, balance, g to mol, mol to g… • Notice that for the whole chapter, I’ll give you info on one thing,you’ll give me info on another • but the only place they can meet is the balanced equation, the only language they speak is “molese”**9.1 Information Given by Chemical Equations**• The cook is a chemist! • Cooking is: mixing ingredients (reactants) in a recipe (equation) to get a certain amount of food (product) • Importance here? that info in a recipe can be used proportionally • double amts? double product!**The chemist is a cook!**• A 2:1mole ratio from a balanced equation is just that - as long as the two substances are in 2:1number ratio (not mass ratio!) it doesn’t matter how much you use**9.2 Mole-Mole Relationships**Remember! The most common number ratio in chemistry is the mole ratio!**Example**How many moles of KClO3 must decompose to produce 9 moles of oxygen gas? (The other product is KCl.) 1) As always! First, get a balanced equation! 2KClO3 2KCl + 3O2 2) What is the mol ratio between KClO3 & O2?**Example (cont’d)**• Remember this? 9 mol O2 2 mol KClO3 = 6 mol KClO3 3 mol O2 from balanced equation! Given**Example**• When 5 mols sodium react with oxygen gas, how many mols of Na2O are formed? 4Na + O2 2Na2O 5 mol Na 2 mol Na2O = 2.50 mol Na2O 4 mol Na**Example**• For this unbalanced equation, how many mols of NO do you need to make 7.5 mol H2O? NO + H2 N2 + H2O • 2NO + 2H2 N2 + 2H2O • it’s a 1:1 ratio! • you need 7.5 mol!**Example**• For this unbalanced equation, how many mols of SiH4 do you need to use with 4.2 mol NH3? SiH4 + NH3 Si3N4 + H2 • 3SiH4 + 4NH3 Si3N4 + 12H2 • it’s a 3:4 ratio! • you need 3.15 mol!**9.3 Mass Calculations**• The study of the amount of substances consumed/produced = stoichiometry • Remember that chemical equations tell you number ratios, not mass amounts! • Your 3 friends now: 1) balanced chemical equations 2) mole ratios 3) math relationships**We’ll start off simple with a problem in which I give you**moles of one but you are asked about massof another… • But remember:the only place they can meet is the balanced equation, the only language they speak is molese**CaC2 and water get together to make acetylene, C2H2, and**calcium hydroxide. How many grams of water do you need to make 1.55 mol C2H2? • Balanced equation!!! CaC2 + 2H2O C2H2 + Ca(OH)2 g mol H2O 1.55 mol C2H2**Example cont’d**• They can only talk mole language from the balanced equation! 1.55 mol C2H2 2 mol H2O 18.0 g H2O 1 mol C2H2 1 mol H2O = 55.8 g H2O mol bridge**Example**• How many grams of potassium chlorate (KClO3) must decompose to produce KCl and 1.45 mol oxygen gas? • First! balance the equation!!! 2KClO3 2KCl + 3O2**g**2KClO3 2KCl + 3O2 mol KClO3 1.45 mol**Example cont’d**• They can only talk mol language from the balanced equation! 1.45 mol O2 2 mol KClO3 122.6 g KClO3 3 mol O2 1 mol KClO3 = 119 g KClO3 mol bridge**Example**• How many moles of copper must react with AgNO3 to produce 5.5 g silver and copper(II) nitrate? • First! balance the equation!!! Cu + 2AgNO3 2Ag + Cu(NO3)2**example cont’d**Cu + 2AgNO3 2Ag + Cu(NO3)2 5.5g mol Ag mol Cu**Example cont’d**• They can only talk mol language from the balanced equation! 5.5 g Ag 1 mol Cu 1 mol Ag 107.9 g Ag 2 mol Ag = 0.025 mol Cu mol bridge**This is for big people only; here I give you mass of one you**tell me mass of the other • key? start and end w/ grams;but must cross thru mol bridge!!! • Basic method? (gramsAmolsA )(molsBgramsB ) PT PT BEq**example**• How many grams of oxygen gas are required to completely react with 14.6 gof solid sodium to form sodium oxide, Na2O?**g**14.6 g 4Na + O2 2Na2O mol O2 mol Na notice that there are always non-players. but they are needed for balancing the equation!**mol bridge**example cont’d • remember: they can only talk mol language from the balanced equation! 14.6 g Na 1 mol Na 1 mol O2 32.0 g O2 23.0 g Na 4 mol Na 1 mol O2 = 5.08 g O2**example**• When 20.4 grams of sodium metal are mixed with chlorine gas, are 52.0 g of sodium chloride produced?**g**20.4 g 2Na + Cl2 2NaCl mol NaCl mol Na**20.4 g Na**2mol NaCl 1 mol Na 58.5 g NaCl 23.0 g Na 2 mol Na 1 mol NaCl = 51.9 g NaCl mol bridge example cont’d • They can only talk mol language from the balanced equation!**example**• Limestone, CaCO3, is heated to produce lime, calcium oxide, CaO, and carbon dioxide. How much limestone is required to produce 10.0 g of lime?**g**10.0 g CaCO3 CaO + CO2 mol CaO mol CaCO3**10. g CaO**1mol CaCO3 1 mol CaO 100.1g CaCO3 56.1 g CaO 1 mol CaO 1 mol CaCO3 = 17.8 g CaCO3 mol bridge example cont’d • They can only talk mol language from the balanced equation!**summary**gA gB PT PT molA molB BE**9.5 Mass Calculations:Comparing Two Reactions**• we can use all this to compare; for example, which store-bought antacid is more effective by weight… • sodium bicarbonate and magnesium hydroxide are both used as antacids • if you have 1.00 g of both which will eat the most acid?**1.00 g NaHCO3**1mol NaHCO3 1 mol HCl 84.01 g NaHCO3 1 mol NaHCO3 mol bridge = 1.19 x 10-2 mol 1) sodium bicarb? NaHCO3 + HCl NaCl + H2O + CO2**2) magnesium hydroxide?**Mg(OH)2 + 2HCl MgCl2 + 2H2O 1.00 g Mg(OH)2 1 mol Mg(OH)2 2 mol HCl 58.33 g Mg(OH)2 1 mol Mg(OH)2 mol bridge = 3.42 x 10-2 mol**Sorry, but equations Do Not describe:**1) Exact conditions * Temperature * Pressure * Volume 2) What the atoms are doing (e.g. not all of them react)**9.6 The Concept ofLimiting Reactants**• Just as if you were to make 100 sandwiches there will be something left over, • Impossible for every atom/molecule to react, so… • must add one reactant in there in excess (xs) to react as much of the other as possible**what is the limiting ingredient here?i.e. what will stop you**from making more sandwiches?**how many sets of 1 pliers, 2 screwdrivers, and 1 hammer can**you make, and what is left over? the “limiting reactant” here were hammers**Mg reacts w/ O2 in 2:1 ratio**• So the 6 Mg on left only need 3 O2’s; but there are 7 O2’s! • what happens to the remaining oxygens???**4 O2’s will not play, but they assure that (essentially)**all Mg’s will be found and destroyed! • Mg’s will run out first; they limit the amount of product that can be made; Mg is the limiting reactant**9.6 EOCs**• 29**9.7 Calculations Involving a Limiting Reactant**• If reactants in a reaction are not present in their mol ratios, one will be used up before the other = limiting reactant; the other one = excess • but you can still predict how much prodruct will be made; yippee! • [hint: limiting reactant probs betray themselves in asking how much product you can get when they give amts of 2 reactants]**You must determine which of the 2 is limiting and finish the**prob w/ that one only, like this: 1) change amts of both reactants (masses) to mols (old stuff) 2) check mol ratios for which one will be used up first (limiting reactant),(new thing!) 3) use that one only to finish prob (forget about the xs reactant)(old stuff)**example**We’ll start simple: • 1.21 mol zinc are added to 2.65 mol HCl. Zinc chloride and hydrogen gas are formed. Which reactant is in XS? Calculate the amt of mols of zinc chloride produced. • START W/ BALANCED EQUATION!**example cont’d**Zn + 2HCl ZnCl2 + H2 • the Zn:HCl is 1:2 • Only one reactant can take us to product (= LR) • Will it be Zn or HCl???**example cont’d**Zn + 2HCl ZnCl2 + H2 • One way is to pretend you don’t know about one of the reactants, e.g... • If there are 1.21 mol Zn, how many mols of HCl would you need? • the Zn:HCl is 1:2, so 1.21 mol Zn need twice as much HCl = 2.42 mol HCl**example cont’d**Zn + 2HCl ZnCl2 + H2 • Do we have 2.42 mol HCl??? • YESSIREEBOB!, and more! (2.65 mol) • We have plenty of HCl (= XS) :) • but we will run out of Zn (= LR) :( • Zn will take us to products!...

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