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Chapter 9

Chapter 9. Stoichiometry. 1) Mole-Mole calc. 2) Mass-Mass calc. 3) Mass-Volume calc. 4) Limiting Reagent 5) Percent Yield. Mole Concept in Chemical Equations. I) Mole-Mole Calculations Al + O 2 Al 2 O 3.

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Chapter 9

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  1. Chapter 9 Stoichiometry

  2. 1) Mole-Mole calc. • 2) Mass-Mass calc. • 3) Mass-Volume calc. • 4) Limiting Reagent • 5) Percent Yield Mole Concept in Chemical Equations

  3. I) Mole-Mole Calculations Al + O2 Al2O3 Ex) How many moles of aluminum are needed to form 2.3 moles of Al2O3? step 1: balance the equation Al + O2 Al2O3 4 3 2

  4. ? X • step 2: identify possible mole ratios 4 Al + O2 Al2O3 3 2 4 mol Al 2 mol Al2O3 2 mol Al2O3 4 mol Al use: or step 3:calculate looking for Al so it goes on top 4 mol Al 2 mol Al2O3 2.3 mol Al2O3 =4.6 mol Al units cancel labels cancel

  5. Ex) How many moles of Al2O3 are formed when 17.2 moles of O2 react with aluminum? 4 3 2 Al + O2 Al2O3 3 mol O2 2 mol Al2O3 2 mol Al2O3 3 mol O2 use mole ratio: or calculate: looking for Al2O3 so it goes on top 2 mol Al2O3 3 mol O2 17.2 mol O2 =11.5 mol Al2O3 do CW #1: prob # 7

  6. Mole-Mole Calculation Practice 1) Calculate the moles of water that can be produced when 0.35 moles of oxygen gas reacts with hydrogen gas. balanced equation: 2 H2 + O2 2 H2O 2 mol H2O 1 mol O2 use mole ratio: looking for H2Oso it goes on top calculate: 2 mol H2O 1 mol O2 0.35 mol O2 =0.70 mol H2O

  7. 2) How many moles of chlorine gas will be required to react with iron to produce 14 moles of iron (III) chloride? balanced equation: 2 Fe + 3 Cl2 2 FeCl3 3 mol Cl2 2mol FeCl3 use mole ratio: looking for Cl2 so it goes on top calculate: 14 mol FeCl3 3 mol Cl2 2 mol FeCl3 = 21 mol Cl2

  8. Stoichiometry Flow Chart unknown mass known mass known moles mole ratio unknown moles unknown liters known liters

  9. II) Mass-Mass Calculations Ex) How many grams of ammonia (NH3) is produced when 5.40 g of hydrogen reacts with nitrogen? step 1: write a balanced equation N2 + 3 H2 2 NH3 step 2: convert grams of known amount to moles 5.40 g H2 mole H2 2.02 g ~

  10. step 3: convert moles of known amount to moles of unknown amount using the correct mole ratio: 2 mol NH3 3 mol H2 ~ ~ step 4: convert unknown moles to grams: ~ 17.0 g NH3 mole NH3 = All in one step: 5.40 g H2 mol H2 2.0 g H2 2 mol NH3 3 mol H2 17.0 g NH3 mol NH3 = = 30.6 g NH3

  11. III) Mass -Volume Calculations Ex) Hydrazine, N2H4, is used as a rocket fuel. It reacts with oxygen to form nitrogen & water. How many liters of nitrogen form when 1.0 kg of hydrazine reacts? N2H4 + O2 N2 + H2O 2 1000g N2H4 mol N2H4 32.0 g N2H4 1 mol N2 1 mol N2H4 22.4 L N2 mol N2 = 700 L N2

  12. Limiting & Excess Reagent (reactant) -The limiting reactant is not present in sufficient quantity to react with all other reactants. -The reaction stops when the limiting reactant is completely consumed. -Any remaining reactants are considered excess reactants. -The amount of product formed is determined by the limiting reactant.

  13. Ex) Determine the limiting reagent for the reaction of 0.333 mol of aluminum reacting with 0.249 mol of sulfur to form aluminum sulfide. Solution:- determine the amount (in moles) of each product formed for each reactant - the reactant producing the lesser amount of product is the limiting reagent

  14. 2 Al + 3 S Al2S3 for aluminum: 1 mol Al2S3 2 mol Al 0.333 mol Al = 0.167 mol Al2S3 for sulfur: 1 mol Al2S3 3 mol S 0.249 mol S = 0.0830 mol Al2S3 There are less moles of Al2S3 formed when the given amount of sulfur was calculated so sulfur is the limiting reagent.

  15. Ex) MgCl2 is produced when 50.6 g sample of Mg(OH)2 is reacted with 45.0 g HCl. Mg(OH)2 + 2 HCl MgCl2 + 2 H2O Determine the limiting reagent. 50.6g Mg(OH)2 mol Mg(OH)2 58.3g Mg(OH)2 1mol MgCl2 1 mol Mg(OH)2 = 0.868 mol MgCl2 45.0 g HCl mol HCl 36.5g HCl 1mol MgCl2 2 mol HCl = 0.616 mol MgCl2 There are less moles of HCl, so it is the limiting reagent.

  16. Percent Yield actual yield theoretical yield Percent yield = x 100 % ex) Calcium carbonate decomposes into CaO and CO2 when heated. The theoretical yield was calculated to be 13.9 g of CaO. If the actual yield was 13.1 g, what was the percent yield of CaO? 13.1 g 13.9 g % yield = x 100% =94.2% CaO

  17. Ex) 45.8 g of potassium carbonate,K2CO3, are reacted with hydrochloric acid. 46.3 g of potassium chloride, KCl, along with water and carbon dioxide are produced. a) Write the balanced equation for the reaction. K2CO3 + 2 HCl 2 KCl + H2O + CO2

  18. b) Calculate the theoretical yield of KCl. 45.8g K2CO3 mol K2CO3 138g K2CO3 2 mol KCl 1 mol K2CO3 74.5g KCl mol KCl = 49.5 g KCl c) Calculate the percent yield of KCl. 46.3 g KCl 49.5 g KCl % yield = x 100 % = 93.5 %

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