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Stoichiometry

Stoichiometry. 6 mol H 2 O . 4 mol NO . 1 1 .33 mol NO. 0.21 1 2 mol H 2 O. x . x . = . = . 6 mol H 2 O. 5 mol O 2. Stoichiometry questions (1). Consider : 4NH 3 + 5O 2  6H 2 O + 4NO How many moles of H 2 O are produced if 0.176 mol of O 2 are used?

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Stoichiometry

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  1. Stoichiometry

  2. 6 mol H2O 4 mol NO 11.33 mol NO 0.2112 mol H2O x x = = 6 mol H2O 5 mol O2 Stoichiometry questions (1) Consider : 4NH3 + 5O2 6H2O + 4NO • How many moles of H2O are produced if 0.176 mol of O2 are used? • How many moles of NO are produced in the reaction if 17 mol of H2O are also produced? # mol H2O= 0.176 mol O2 # mol NO= 17 mol H2O Notice that a correctly balanced equation is essential to get the right answer

  3. 6 mol H2O 32 g O2 18.02 g H2O 5 mol O2 8 g O2 51.4 g H2O x x x x = = 6 mol H2O 1 mol H2O 4 mol NH3 1 mol O2 Stoichiometry questions (2) Consider : 4NH3 + 5O2 6H2O + 4NO • How many grams of H2O are produced if 1.9 mol of NH3 are combined with excess oxygen? • How many grams of O2 are required to produce 0.3 mol of H2O? # g H2O= 1.9 mol NH3 # g O2= 0.3 mol H2O

  4. 4 mol NO 30.01 g NO 1 mol O2 x x x 5 mol O2 1 mol NO 32 g O2 = 9.0 g NO Stoichiometry questions (3) Consider : 4NH3 + 5O2 6H2O + 4NO • How many grams of NO is produced if 12 g of O2 is combined with excess ammonia? # g NO= 12 g O2

  5. DO NOW- Balance • Al + HCl  AlCl3 + H2 • N2 + H2 NH3 • Al + O2 Al2O3 • C3H8 + O2 CO2 + H2O • 2Al + 6HCl  2AlCl3 + 3H2 • N2 + 3H2 2NH3 • 4Al + 3O2 2 Al2O3 • C3H8 + 5O2 3CO2 + 4H2O

  6. = = 35.3 g NO 22.5 g NO 1 mol NH3 1 mol O2 4 mol NO 4 mol NO 30.0 g NO 30.0 g NO x x x x x x 32.0 g O2 17.0 g NH3 4 mol NH3 5 mol O2 1 mol NO 1 mol NO Limiting Reagents: • Do two separate calculations using both given quantities. The smaller answer is correct. Q - How many g NO are produced if 20 g NH3 is burned in 30 g O2? 4NH3+5O26H2O+4NO # g NO= 20 g NH3 30 g O2

  7. 1 mol AgNO3 1 mol MgCl2 2 mol AgCl 2 mol AgCl x x x x 1 mol MgCl2 2 mol AgNO3 169.88 g AgNO3 95.21 g MgCl2 = = 75.25 g AgCl 57.36 g AgCl 143.3 g AgCl 143.3 g AgCl x x 1 mol AgCl 1 mol AgCl Limiting Reagents MgCl2 + 2AgNO3 Mg(NO3)2 + 2AgCl If 25 g magnesium chloride was added to 68 g silvernitrate,whatmassofAgClwillbeproduced? # g AgCl= 25 g MgCl2 # g AgCl= 68 g AgNO3 For more lessons, visit www.chalkbored.com

  8. 2Al + 6HCl  2AlCl3 + 3H2 If 25 g of aluminum was added to 90 g of HCl, what mass of H2 will be produced? • N2 + 3H2 2NH3: If you have 20 g of N2 and 5.0 g of H2, which is the limiting reagent? • What mass of aluminum oxide is formed when 10.0 g of Al is burned in 20.0 g of O2? • When C3H8 burns in oxygen, CO2 and H2O are produced. If 15.0 g of C3H8 reacts with 60.0 g of O2, how much CO2 is produced? Practice questions-LR

  9. 1 mol HCl 1 mol Al 3 mol H2 3 mol H2 x x x x 2 mol Al 6 mol HCl 36.5 g HCl 27.0 g Al = 2.47 g H2 = 2.78 g H2 2.0 g H2 2.0 g H2 x x 1 mol H2 1 mol H2 Question 1 2Al + 6HCl  2AlCl3 + 3H2 If 25 g aluminum was added to 90 g HCl,whatmassofH2willbeproduced? # g H2= 25 g Al # g H2 = 90 g HCl

  10. 1 mol H2 1 mol N2 2 mol NH3 2 mol NH3 x x x x 1 mol N2 3 mol H2 2.0 g H2 28.0 g N2 = 24.3 g NH3 = 28.3 g NH3 17.0 g NH3 17.0 g NH3 x x 1 mol NH3 1 mol NH3 Question 2 N2 + 3H2 2NH3 If you have 20 g of N2 and 5.0 g of H2, which is the limiting reagent? # g NH3= 20 g N2 # g NH3 = 5.0 g H2 N2 is the limiting reagent

  11. 1 mol O2 1 mol Al 2 mol Al2O3 2 mol Al2O3 x x x x 4 mol Al 3 mol O2 32.0 g O2 27.0 g Al = 42.5 g Al2O3 = 18.9 g Al2O3 102.0 g Al2O3 102.0 g Al2O3 x x 1 mol Al2O3 1 mol Al2O3 Question 3 4Al + 3O2 2 Al2O3 What mass of aluminum oxide is formed when 10.0 g of Al is burned in 20.0 g of O2? # g Al2O3= 10.0 g Al # g Al2O3= 20.0 g O2

  12. 1 mol O2 1 mol C3H8 3 mol CO2 3 mol CO2 x x x x 1 mol C3H8 5 mol O2 32.0 g O2 44.0 g C3H8 = 45.0 g CO2 = 49.5 g CO2 44.0 g CO2 44.0 g CO2 x x 1 mol CO2 1 mol CO2 Question 4 C3H8 + 5O2 3CO2 + 4H2O When C3H8 burns in oxygen, CO2 and H2O are produced. If 15.0 g of C3H8 reacts with 60.0 g of O2, how much CO2 is produced? # g CO2= 15.0 g C3H8 # g CO2= 60.0 g O2 5. Limiting reagent questions give values for two or more reagents (not just one)

  13. 1 mol H2 actual 2 mol H2O 138 g H2O x = x 2.02 g H2 theoretical 143 g H2O 2 mol H2 18.02 g H2O x 1 mol H2O = = 143 g 96.7% Percent Yield Q - What is the % yield of H2O if 138 g H2O is produced from 16 g H2 and excess O2? Step 1: write the balanced chemical equation 2H2 + O2 2H2O Step 2: determine actual and theoretical yield. Actual is given, theoretical is calculated: # g H2O= 16 g H2 Step 3: Calculate % yield % yield = x 100 x 100

  14. actual 2 mol NH3 40.5 g NH3 x = theoretical 3 mol H2 227 g NH3 17.04 g NH3 x 1 mol NH3 = = 17.8% 227 g Practice problem Q - What is the % yield of NH3 if 40.5 g NH3 is produced from 20.0 mol H2 and excess N2? Step 1: write the balanced chemical equation N2 + 3H2 2NH3 Step 2: determine actual and theoretical yield. Actual is given, theoretical is calculated: # g NH3= 20.0 mol H2 Step 3: Calculate % yield % yield = x 100 x 100

  15. 1 mol O2 actual 1 mol H2 2 mol H2O 2 mol H2O 58 g H2O x x x x = 32 g O2 2.02 g H2 theoretical 1 mol O2 2 mol H2 62.4 g H2O 18.02 g H2O 18.02 g H2O x x 1 mol H2O 1 mol H2O = = 92.9% 68 g = 62.4 g Challenging question 2H2 + O2 2H2O What is the % yield of H2O if 58 g H2O are produced by combining 60 g O2 and 7.0 g H2? Hint: determine limiting reagent first # g H2O= 60 g O2 # g H2O= 7.0 g H2 % yield = x 100 x 100

  16. More Percent Yield Questions The electrolysis of water forms H2 and O2. 2H2O  2H2 + O2 What is the % yield of O2 if 12.3 g of O2 is produced from the decomposition of 14.0 g H2O? • 107 g of oxygen is produced by heating 300 grams of potassium chlorate. Calculate % yield. 2KClO3 2KCI + 3O2 • What is the % yield of ferrous sulphide if 3.00 moles of Fe reacts with excess sulfur to produce 220 grams of ferrous sulphide? Fe + S  FeS

  17. More Percent Yield Questions • Iron pyrites (FeS2) reacts with oxygen according to the following equation: 4FeS2 + 11O2 2Fe2O3 + 8SO2 If 300 g of iron pyrites is burned in 200 g of O2, 143 grams of ferric oxide is produced. What is the percent yield of ferric oxide? • 70 grams of manganese dioxide is mixed with 3.5 moles of hydrochloric acid. How many grams of Cl2 will be produced from this reaction if the % yield for the process is 42%? MnO2 + 4HCI  MnCl2 + 2H2O + Cl2

  18. 1 mol H2O actual 1 mol O2 12.3 g O2 x = x 18.02 g H2O theoretical 12.43 g O2 2 mol H2O 32 g O2 x 1 mol O2 = = 12.43 g 98.9% Q1 • The electrolysis of water forms H2 & O2. 2H2O  2H2 + O2 Give the percent yield of O2 if 12.3 g O2 is produced from the decomp. of 14 g H2O? • Actual yield is given: 12.3 g O2 • Next, calculate theoretical yield # g O2= 14.0 g H2O Finally, calculate % yield % yield = x 100% x 100%

  19. 1 mol KClO3 actual 3 mol O2 107 g O2 x = x 122.55 g KClO3 theoretical 117.5 g O2 2 mol KClO3 32 g O2 x 1 mol O2 = = 117.5 g 91.1% Q2 • 107 g of oxygen is produced by heating 300 grams of potassium chlorate. 2KClO3 2KCI + 3O2 • Actual yield is given: 107 g O2 • Next, calculate theoretical yield # g O2= 300 g KClO3 Finally, calculate % yield % yield = x 100% x 100%

  20. actual 1 mol FeS 220 g O2 x = theoretical 1 mol Fe 263.7 g O2 87.91 g FeS x 1 mol FeS = = 83.4% 263.7 g Q3 • What is % yield of ferrous sulfide if 3 mol Fe produce 220 grams of ferrous sulfide? Fe + S  FeS • Actual yield is given: 220 g FeS • Next, calculate theoretical yield # g FeS= 3.00 mol Fe Finally, calculate % yield % yield = x 100 x 100

  21. actual 143 g Fe2O3 = theoretical 181.48 g Fe2O3 = = = 181.48 g Fe2O3 78.8% 199.7 g Fe2O3 1 mol O2 1 mol FeS2 2 mol Fe2O3 2 mol Fe2O3 159.7 g Fe2O3 159.7 g Fe2O3 x x x x x x 32 g O2 119.97 g FeS2 4 mol FeS2 11 mol O2 1 mol Fe2O3 1 mol Fe2O3 • 4FeS2 + 11O2 2Fe2O3 + 8SO2 If 300 g of FeS2 is burned in 200 g of O2, 143 g Fe2O3 results. % yield Fe2O3? First, determine limiting reagent # g Fe2O3= 300 g FeS2 200 g O2 % yield = x 100% x 100

  22. actual 42 x 57.08 g Cl2 x g Cl2 = = theoretical 100 57.08 g Cl2 = = = = 42% 57.08 g Cl2 24 g Cl2 62.13 g Cl2 1 mol MnO2 1 mol Cl2 1 mol Cl2 71 g Cl2 70.9 g Cl2 x x x x x 86.94 g MnO2 4 mol HCl 1 mol MnO2 1 mol Cl2 1 mol Cl2 # g Cl2= • 70 g of MnO2 + 3.5 mol HCl gives a 42% yield. How many g of Cl2 is produced? MnO2 + 4HCI  MnCl2 + 2H2O + Cl2 70 g MnO2 3.5 mol HCl % yield = x 100% x 100 x g Cl2 For more lessons, visit www.chalkbored.com

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