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Chapter 10

Chapter 10. Q1. Calculate the molar mass of the following compounds: NaOH - 40g/mol AuCl 3 – 303.32g/mol (NH 4 ) 2 SO 4 – 132.16g/mol. Q2. Convert 3.4 moles of NH 4 into grams of NH 4. 3.4 mole of NH 4. 18.04g of NH 4. 61g. =. 1 mol of NH 4. Q3.

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Chapter 10

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  1. Chapter 10

  2. Q1 • Calculate the molar mass of the following compounds: • NaOH - 40g/mol • AuCl3 – 303.32g/mol • (NH4)2SO4 – 132.16g/mol

  3. Q2 • Convert 3.4 moles of NH4 into grams of NH4. 3.4 mole of NH4 18.04g of NH4 61g = 1 mol of NH4

  4. Q3 Convert 4.2 grams of Ca(NO3)2 into moles of Ca(NO3)2. 4.2g of Ca(NO3)2 1mol of Ca(NO3)2 0.026mol = 164.1g of Ca(NO3)2

  5. Q4 Convert 7.8 L of N2 into moles. 1mol of N2 7.8L of N2 0.35 mol = 22.4L of N2

  6. Q5 Convert 3.5 mol of H2 into liters 22.4L of H2 3.5 mol of H2 78 L of H2 = 1 mol of H2

  7. Q6 Convert 1.4 moles of CO2 into molecules of CO2. 6.02 x 1023 molecules of CO2 1.4 mol of CO2 8.43 x 1023 = 1 mol of CO2

  8. Q7 Convert 2.3 X 1025 atoms of Mg(NO3)2 into grams of Mg(NO3)2. 2.3 x 1025 molecules of Mg(NO3)2 1 mol of Mg(NO3)2 148.31g of Mg(NO3)2 5.7 X 103g = 6.02 x 1023 molecules of Mg(NO3)2 1 mol of Mg(NO3)2

  9. Q8 What is the percent composition of sulfur in Al2(SO4)3? Part (3 x molar mass of S) X 100% = % Comp Whole (molar mass of Al2(SO4)3) 96.18g = 28.1% S X 100% 342.15g

  10. Q9 What is the percent composition of carbon in pentane, C5H12? Part (5 x molar mass of C) X 100% = % Comp Whole (molar mass of C5H12) 60.05g = 83.23% C X 100% 72.15g

  11. q10

  12. Chapter 12

  13. (Ch 12) Q1Mole-Mole Problem: According to the following equation, how many moles of Al react with 5.2 moles of H2SO4? 2Al + 3H2SO4 Al2(SO4)3 + 3H2 5.2 moles of H2SO4 2 mol of Al 3.5 mole = 3 moles of H2SO4

  14. (Ch 12) Q2Mole-Mole Problem: According to the following equation, how many moles of oxygen are needed to form 3.7 mol Al2O3? 4Al + 3O2 2Al2O3 3.7 moles of Al2O3 3 mol of O2 5.6 mole = 2 moles of Al2O3

  15. (Ch 12) Q3Mass-Mole Problem: According to the equation in problem #2, how many moles of Al2O3 are formed when 7.8 grams of aluminum completely reacts with oxygen? 4Al + 3O2 2Al2O3 2 mol of Al2O3 7.8g of Al 1 mol of Al 0.14mole = 4 mol of Al 26.98g of Al

  16. (Ch 12) Q4Mass-Mole-Mole-Mass Problem: According to the following equation, how many grams of CO2 would be produced if 45g of C2H2 completely reacted with oxygen? 2C2H2 + 5O2 4CO2 + 2H2O 4 mol CO2 44.0g of CO2 45g of C2H2 1 mol C2H2 152.1g CO2 = 2 mol C2H2 26.0g of C2H2 1 mol CO2

  17. (Ch 12) Q5a/b/c In the equation below 3.4g of Al reacts with 4.2g of O2 to form aluminum oxide. a.) What is the limiting reactant? Limiting Reagent 4Al + 3O2 2Al2O3 2 mol Al2O3 3.4g of Al Theoretical Yield 1 mol Al 101.9g Al2O3 6.4g Al2O3 = 4 mol Al 27.0g of Al 1 mol Al2O3 Excess Reagent 2 mol Al2O3 4.2g of O2 1 mol O2 101.9g Al2O3 8.9g Al2O3 = 32.0g of O2 1 mol Al2O3 3 mol O2

  18. (Ch 12) Q5d d.) How much of the excess reactant is left over? 4Al + 3O2 2Al2O3 3 mol O2 6.4g of Al2O3 1 mol Al2O3 32.0g O2 3.02g O2 = 4 mol Al 1 mol O2 101.9g Al2O3 Excess Reactant (O2) = 4.2-3.02g = 1.18g

  19. (Ch 12) Q5e e) If the actual yield of aluminum oxide is 1.24g Al2O3 what is the percent yield? 4Al + 3O2 2Al2O3 Actual yield (grams) X 100% = % Yield Theoretical yield (grams) 1.24g of Al2O3 = 19.3% X 100% 6.4g of Al2O3

  20. (Ch 12) Q6 Sodium chloride decomposes through electrolysis by the below. What is the percent yield of sodium produced if 45g of NaCl reacts and produces an actual yield of 6.2g of Na? 2NaCl  2Na + Cl2 2 mol Na 45g of NaCl Theoretical Yield 1 mol NaCl 23.0g Na 17.67g of Na = 58.44g of NaCl 1 mol Na 2 mol NaCl

  21. (Ch 12) Q6 Sodium chloride decomposes through electrolysis by the below. What is the percent yield of sodium produced if 45g of NaCl reacts and produces an actual yield of 6.2g of Na? 2NaCl  2Na + Cl2 6.2g of Na (actual yield) = 35% X 100% 17.67g Na (Theoretical yield)

  22. (Ch 12) Q7 The combustion of propane, C3H8 is shown in the equation below. Identify the limiting reactant and the volume of CO2 formed when 6 L of C3H8 reacts with 12 L of O2 to produce CO2 gas and H2O vapor at STP. C3H8(g) + 5O2(g)  3CO2(g) + 4H2O (g) 6L of C3H8 3L of CO2 18L of CO2 = 1L of C3H8 Limiting Reactant 12L of O2 3L of CO2 7.2L of CO2 = 5L of O2

  23. (Ch 12) Q8 From problem #7, how much of excess reactant is left unreacted? C3H8(g) + 5O2(g)  3CO2(g) + 4H2O (g) 1L of C3H8 7.2L of CO2 2.4L of C3H8 needed = 3L of CO2 Excess Reactant (C3H8) = 6-2.4 = 3.6L

  24. (Ch 12) Q9 From problem #7, if the actual yield was 6.9 L of CO2, what is the percent yield? C3H8(g) + 5O2(g)  3CO2(g) + 4H2O (g) 6.9L of CO2(actual yield) = 95.8% X 100% 7.2L CO2(Theoretical yield)

  25. (Ch 12) Q10a In the reaction below, 5.50g of Mg reacts with 3.25g of O2 to produce MgO. a) Balance this equation: 2 Mg + 1 O2 2 MgO

  26. (Ch 12) Q10b/c/d In the reaction below, 5.50g of Mg reacts with 3.25g of O2 to produce MgO. b) Which is the theoretical yield of MgO? 2 Mg + 1 O2 2 MgO Excess Reactant 2 mol MgO 5.5g of Mg 1 mol Mg 40.3g MgO 9.12g MgO = 2 mol Mg 24.3g of Mg 1 mol MgO Limiting Reactant 2 mol MgO 3.25g of O2 1 mol O2 40.3g MgO 8.19g MgO = 32.0g of O2 1 mol O2 1 mol MgO

  27. (Ch 12) Q10e In the reaction below, 5.50g of Mg reacts with 3.25g of O2 to produce MgO. e) How much of the excess reactant is left unreacted? 2 Mg + 1 O2 2 MgO 2 mol Mg 8.19g MgO 1 mol MgO 24.3g Mg 4.94g Mg = 40.3g MgO 2 mol MgO 1 mol Mg Excess Reactant (Mg) = 5.50-4.94= 0.56g

  28. (Ch 12) Q10f In the reaction below, 5.50g of Mg reacts with 3.25g of O2 to produce MgO. f) Calculate the percent yield of MgO if the actual yield was 6.80g MgO. 2 Mg + 1 O2 2 MgO 6.8g of MgO (actual yield) = 83.0% X 100% 8.19g MgO (Theoretical yield)

  29. Chapter 14

  30. Ch14 Q5 P1V1 = P2V2 P1 V1 A gas has a volume of 50cm3 at a pressure of 200 mm Hg. If the pressure is changed to 190 mm Hg, what is the new volume of that sample of gas? P2 P1V1 200 x 50 P1V1= P2V2 = V2 = P2 190 V2 = 53cm3

  31. Ch14 WP Q6V1/T1 = V2/T2 V1 T1 If 400mL of nitrogen is collected at 30oC, what will the new volume be if the temperature is increased to 50oC? T2 V1 V2 400 x 323 V1T2 = = V2 = T1 T2 T1 303 V1 = 426mL

  32. Ch14 Q7P1/T1= P2/T2 P1 T1 A cylinder of carbon dioxide left outside has a pressure of 3.5 atm and a temperature of 50oC. If the cylinder is taken indoors and cooled to 25oC, what will the new pressure be? T2 P1 P2 3.5 x 298 P1T2 = = P2 = T1 T2 T1 323 P2 = 3.2 atm

  33. Ch14 Q8 P1V1 = P2V2 P1 V1 A gas has a volume of 125 mL at a pressure of 2 atm. If the pressure is changed to 7.2 atm, what is the new volume of that sample of gas? P2 P1V1 2 x 125 P1V1= P2V2 = V2 = P2 7.2 V2 = 35mL

  34. Ch14 Q9P1/T1= P2/T2 T1 P1 To what temperature must a sample of oxygen gas at 350 K and pressure of 4.2 atm be cooled to so the final pressure is 1.25 atm? P2 P1 P2 1.25 x 350 P2T1 = T2 = = T1 T2 P1 4.2 T2 = 104K

  35. Ch14 Q10V1/T1 = V2/T2 T1 V1 A gas has a volume of 20 L at a temperature of 350 K. What will the volume of the gas occupy if the Kelvin temperature is increased to 423 K? T2 V1 V2 20 x 423 V1T2 = = V2 = T1 T2 T1 350 V2 = 24 L

  36. Chapter 16

  37. Ch16 Q1M = mol/L Calculate the molarity of a 3.4 mol of Na2CO3 dissolved in a 100mL solution? 3.4 mole = 34M 0.100L

  38. Ch16 Q2M = mol/L Calculate the molarity of a solution prepared by dissolving 14.5 g of BaCl2 in 250 mL of solution. 14.5g of BaCl2 1 mol BaCl2 0.069mole of BaCl2 = 208.2g of BaCl2 0.069 mole = 0.028M 0.250L

  39. Ch16 Q3ML = mol Calculate the number of grams of solute needed to make a 150 mL solution of 0.40M H2SO4. 0.4M x 0.150L 0.06mole of H2SO4 = 0.06mole x 89.08g/mole = 5.88g of H2SO4

  40. Ch16 Q4L = mol/M If you have 5.2 g of CaCl2 and you want to make a 2.5 M solution, what volume in liter should CaCl2 be diluted to? 5.2g of CaCl2 1 mol CaCl2 0.047mole of CaCl2 = 110.98g of CaCl2 0.047 mole = 0.019L 2.5M

  41. Ch16 Q5M1V1 = M2V2 M2 You want to make a solution with a final concentration of 2.25M HCl from a stock solution of 6.0M HCl. How many milliliters of the stock solution are required to make a solution with a final volume of 50mL? V2 M1 M2V2 2.25 x 50 V1 M1V1 = M2V2 = = M1 6.0 V1 = 18.75mL

  42. Ch16 Q6M1V1 = M2V2 M2 v2 You want to make a 2.50 L solution of 3.50 M HNO3 from a 9.0M HNO3 solution. How many mL of the 9.0M HNO3 should you use? (1L = 1,000mL) M1 M2V2 3.50 x 2.50 V1 M1V1 = M2V2 = = M1 9.0 V1 = 972mL

  43. Ch16 Q7M1V1 = M2V2 M2 v2 You want to make a 250 mL solution of 0.75 M H2SO4 from a 6.0M H2SO4 solution. How many mL of the 6.0M H2SO4 should you use? M1 M2V2 0.75 x 250 V1 M1V1 = M2V2 = = M1 6.0 V1 = 31.25mL

  44. Ch16 Q8M1V1 = M2V2 M1 How many mL of a 15.8M HNO3 stock solution are required to make a 3.0M HNO3 solution with a final volume of 250mL? M2 V2 M2V2 3.0 x 250 V1 M1V1 = M2V2 = = M1 15.8 V1 = 47.5mL

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