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Chemistry Chapter 12 Review

Chemistry Chapter 12 Review. Remember to bring with you your Chapter 12 foldable, periodic table and a calculator if you wish to use one on the test. What is the amount of product formed when a reaction is carried out in the laboratory called?. Limiting reagent Actual yield Percent yield

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Chemistry Chapter 12 Review

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  1. Chemistry Chapter 12 Review Remember to bring with you your Chapter 12 foldable, periodic table and a calculator if you wish to use one on the test.

  2. What is the amount of product formed when a reaction is carried out in the laboratory called? • Limiting reagent • Actual yield • Percent yield • Theoretical yield

  3. What is the reactant that determines the amount of product that can be formed in a reaction is called? • Excess reagent • Percent yield • Limiting reagent • Theoretical yield

  4. What is the maximum amount of product that can be formed from given amounts of reactants called? • Theoretical yield • Percent yield • Excess reagent • Hypothetical yield

  5. What is the reactant that is not completely used up in the reaction called? • Limiting reagent • Excess reagent • Abundant reagent • Useless reagent

  6. What is the ratio of the actual yield to the theoretical yield called? • Ratio yield • Hypothetical yield • Percent yield • Generic yield

  7. In a chemical reaction, the mass of the products… • Is less than the mass of the reactants • Is greater than the mass of the reactants • Is equal to the mass of the reactants • Has no relationship with the mass of the reactants

  8. The first step in most stoichiometry problems is to convert the given quantities to what? moles

  9. Balance the following reaction:__CO + __O2 __CO2 2 CO + O2 2 CO2

  10. 2 CO + O2 2 CO2What is the ratio of moles of oxygen to moles of CO2? 1:2

  11. 2 CO + O2 2 CO2How many moles of oxygen are needed to react completely with 5.4 moles of CO? 5.4 mol CO (1 mol O2/2 mol CO) = 2.7 mol O2

  12. 2 CO + O2 2 CO2How many liters of CO are required to react completely with 3.6 liters of oxygen? 3.6 L O2(1mol/22.4 L)(2mol CO/ 1 mol O2)(22.4 L/1 mol) 7.2 L CO

  13. 2 CO + O2 2 CO2If there are 11.5 g of CO that reacts how many grams of CO2 are produced? 11.5 g CO ( 1 mol/28.01 g CO)(2 mol CO2/2 mol CO)(44.01 g CO2/1 mol) 18.07 g CO2

  14. 2 CO + O2 2 CO2What is the limiting reagent when 6.2 g O2 reacts with 4.0 g CO? 6.2 g O2 (1 mol/32.0 g) = .19 mol O2 4.0 g CO (1 mol/28.01) = .14 mol CO .19 mol O2 (2 mol CO/1 mol O2) = .38 CO .14 < .38 You don’t have enough CO so CO is the limiting reagent.

  15. 2 CO + O2 2 CO2What is maximum amount of CO2 that can be produced when 6.2 g O2 reacts with 4.0 g CO? 4.0 g CO(1 mol/28.01)(2 mol CO2/2 mol CO)(44.01 g CO2/1 mol) = 6.28 g CO2

  16. 2 CO + O2 2 CO2If you actually did this reaction in a lab and produced 5.8 g CO2 what would your percent yield be? (5.8 g/6.28 g) x 100 = 92%

  17. Balance the following reaction:__Zn + __MoO3 __Mo2O3 + __ZnO 3 Zn + 2 MoO3 Mo2O3 + 3ZnO

  18. 3 Zn + 2 MoO3 Mo2O3 + 3ZnOWhat is the ratio of moles of Zn to moles of MoO3? 3:2

  19. 3 Zn + 2 MoO3 Mo2O3 + 3ZnOHow many moles of Zn are needed to react completely with 5.4 moles of MoO3? 5.4 molMoO3 (3 molZn/2 molMoO3) = 8.1molMoO3

  20. 3 Zn + 2 MoO3 Mo2O3 + 3ZnOHow many liters of Zn are required to react completely with 3.6 liters of MoO3? 3.6 L MoO3(1mol/22.4 L)(3mol Zn/ 2 molMoO3)(22.4 L/1 mol) 5.4L Zn

  21. 3 Zn + 2 MoO3 Mo2O3 + 3ZnOIf there are 11.5 g of Zn that reacts how many grams of ZnOare produced? 11.5 g Zn ( 1 mol/65.39 g Zn)(3molZnO/3 molZn)(81.39 g ZnO/1 mol) 14.31 g ZnO

  22. 3 Zn + 2 MoO3 Mo2O3 + 3ZnOWhat is the limiting reagent when 6.2 g Zn reacts with 4.0 g MoO3? 6.2 g Zn (1 mol/65.39 g) = .09 molZn 4.0 g MoO3(1 mol/143.94) = .03molMoO3 .09 molZn (2 molMoO3/3 molZn) = .06CO .03< .06 You don’t have enough MoO3so MoO3is the limiting reagent.

  23. 3 Zn + 2 MoO3 Mo2O3 + 3ZnOWhat is maximum amount of ZnOthat can be produced when 6.2 g Zn reacts with 4.0 g MoO3? 4.0 g MoO3(1 mol/143.94)(3 molZnO/2 molMoO3)(81.39 g ZnO/1 mol) = 3.39 g ZnO

  24. 3 Zn + 2 MoO3 Mo2O3 + 3ZnOIf you actually did this reaction in a lab and produced 2.82 g ZnOwhat would your percent yield be? (2.82 g/3.39 g) x 100 = 83%

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