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Master Chemistry with Jeopardy: Balancing Formulas and Stoichiometry Challenges

Test your chemistry knowledge with this engaging Jeopardy-style review game! Dive into balancing chemical equations, calculate mole ratios, and explore stoichiometry concepts with a variety of challenging questions. From balancing sulfuric acid reactions to determining percent yields, you'll reinforce your understanding and preparation for chemistry exams. This interactive review is perfect for students looking to sharpen their skills in a fun way. Join now and conquer the chemistry challenges ahead!

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Master Chemistry with Jeopardy: Balancing Formulas and Stoichiometry Challenges

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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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