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Jeopardy

Jeopardy. Chemistry Chapter 11 Review Game **Complete Balancing Formulas Section before moving on to the rest of the questions…. Select a Category. Balancing Formulas. __H 2 SO 4 + __Pb(OH) 4  __Pb(SO 4 ) 2 + __H 2 O. 1 point. Check. Balancing Formulas.

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Jeopardy

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  1. Jeopardy Chemistry Chapter 11 Review Game **Complete Balancing Formulas Section before moving on to the rest of the questions…

  2. Select a Category

  3. Balancing Formulas __H2SO4 + __Pb(OH)4 __Pb(SO4)2 + __H2O 1 point Check

  4. Balancing Formulas _2_H2SO4 + _1_Pb(OH)4 _1_Pb(SO4)2 + _4_H2O 1 point Back to Category Slide

  5. Balancing Formulas __Al + __HCl  __AlCl3 + __H2 2 points Check

  6. Balancing Formulas _2_Al + _6_HCl  _2_AlCl3 + _3_H2 2 points Back to Category Slide

  7. Balancing Formulas __CO + __H2 __C8H18 + __H2O 3 points Check

  8. Balancing Formulas _8_CO + _17_H2 _1_C8H18 + _8_H2O 3 points Back to Category Slide

  9. Balancing Formulas __HClO4 + __P4O10 __H3PO4 + __Cl2O7 4 points Check

  10. Balancing Formulas _12_HClO4 + _1_P4O10 _4_H3PO4 + _6_Cl2O7 4 points Back to Category Slide

  11. Balancing Formulas Ammonium Phosphate reacts with Lead IV Nitrate in a double replacement reaction. 5 points Check

  12. Balancing Formulas _4_(NH4)3PO4 + _3_Pb(NO3)4 _1_Pb3(PO4)4 + _12_NH4NO3 5 points Back to Category Slide

  13. Easy Stoichiometry_2_H2SO4 + _1_Pb(OH)4 _1_Pb(SO4)2 + _4_H2O The mole to mole ratio between lead IV hydroxide and dihydrogen monoxide. 1 point Check

  14. Easy Stoichiometry What is 1 to 4? 1 point Back to Category Slide

  15. Easy Stoichiometry_2_H2SO4 + _1_Pb(OH)4 _1_Pb(SO4)2 + _4_H2O 2.5 moles of sulfuric acid are used with excess reactants. This many moles of lead IV sulfate will be produced. 2 points Check

  16. Easy Stoichiometry What is 1.3 (sf) moles of lead IV sulfate? 2 points Back to Category Slide

  17. Easy Stoichiometry_2_H2SO4 + _1_Pb(OH)4 _1_Pb(SO4)2 + _4_H2O The number of grams of water that are produced when 2.5 moles of lead IV hydroxide are used with excess sulfuric acid. 3 points Check

  18. Easy Stoichiometry What is 180 g of H2O? 3 points Back to Category Slide

  19. Easy Stoichiometry_2_H2SO4 + _1_Pb(OH)4 _1_Pb(SO4)2 + _4_H2O The number of mL of sulfuric acid that are needed to produce 15.7 grams of water. 4 points Check

  20. Easy Stoichiometry What is 9770 mL? 4 points Back to Category Slide

  21. Easy Stoichiometry_2_H2SO4 + _1_Pb(OH)4 _1_Pb(SO4)2 + _4_H2O The number of hydrogen atoms that are found in 4.5 moles of water. 5 points Check

  22. Easy Stoichiometry What is 5.4 x 1024 H atoms? 5 points Back to Category Slide

  23. Hard Stoichiometry_12_HClO4 + _1_P4O10 _4_H3PO4 + _6_Cl2O7 The number of moles of tetraphosphorus decaoxide that are needed to react completely with 1.4 mol of HClO4 (perchloric acid) 1 point Check

  24. Hard Stoichiometry What is 0.12 moles? 1 point Back to Category Slide

  25. Hard Stoichiometry_12_HClO4 + _1_P4O10 _4_H3PO4 + _6_Cl2O7 The number of moles produced of Cl2O7 when 4.5 g P4O10 react with 85 g of HClO4. 2 points Check

  26. Hard Stoichiometry What is 0.095 mol Cl2O7? 2 points Back to Category Slide

  27. Hard Stoichiometry_12_HClO4 + _1_P4O10 _4_H3PO4 + _6_Cl2O7 The number of Cl atoms that are produced when 23.5 g of HClO4 react with 22.5 g of P4O10. 3 points Check

  28. Hard Stoichiometry What is 1.41 x 1023 Cl atoms? 3 points Back to Category Slide

  29. Hard Stoichiometry_12_HClO4 + _1_P4O10 _4_H3PO4 + _6_Cl2O7 The number of oxygen atoms that will be found in products when 2.5 g of both HClO4 and P4O10 react together. 4 points Check

  30. Hard Stoichiometry What is 7.2 x 1022 O atoms? 5 points Back to Category Slide

  31. % Yield Three reasons when % composition can never be greater than 100%. 1 point Check

  32. % Yield What is: 1) impure reactants 2) Competing side reactions 3) Loss of reactants and products during transfer 1 point Back to Category Slide

  33. % Yield_4_(NH4)3PO4 + _3_Pb(NO3)4 _1_Pb3(PO4)4 + _12_NH4NO3 The % composition when 12.5 grams of lead IV nitrate are produced in a lab, but mathematically was supposed to produce 13.4 g. 2 points Check

  34. % Yield What is 93.3% yield? 2 points Back to Category Slide

  35. % Yield Determine the percent yield for the reaction between 2.80 g Al(NO3)3 and excess NaOH if 0.966 g Al(OH)3 is recovered. 3 points Check

  36. % Yield What is 93.8%? 3 points Back to Category Slide

  37. % Yield Determine the percent yield for the reaction between 15.0 g N2 and 15.0 g H2 if 10.5 g NH3 are produced. 4 points Check

  38. % Yield What is 57.7%? 4 points Back to Category Slide

  39. % Yield A piece of copper with a mass of 5.00 g is placed in a solution of silver nitrate containing excess AgNO3. The silver metal produced has a mass of 15.2 g. This is the percent yield for this reaction. 5 points Check

  40. % Yield What is 89.4%? 5 points Back to Category Slide

  41. Congratulations!You have completed the game of Jeopardy.

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