1 / 69

$100

$100. $100. $100. $100. $200. $200. $200. $200. $300. $300. $300. $300. $300. $400. $400. $500. $500. $500. $500. $500. $100. $200. $400. $400. $400. Simple Stoich. Grams to Grams. Limiting Reactants. Review. Sing Me Baby One More Time. Simple Stoich.

aderyn
Télécharger la présentation

$100

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. $100 $100 $100 $100 $200 $200 $200 $200 $300 $300 $300 $300 $300 $400 $400 $500 $500 $500 $500 $500 $100 $200 $400 $400 $400

  2. Simple Stoich

  3. Grams to Grams

  4. Limiting Reactants

  5. Review

  6. Sing Me Baby One More Time

  7. Simple Stoich Limiting Reactant Grams to Grams Sing Me Baby One More Time Review $100 $100 $100 $100 $100 $200 $200 $200 $200 $200 $300 $300 $300 $300 $300 $400 $400 $400 $400 $400 $500 $500 $500 $500 $500

  8. CATEGORY 1 - $100 This many moles of Aluminum Oxide would be produced from 12 moles of Aluminum in the equation: 4Al + 3O2 2Al2O3

  9. CATEGORY 1 - $200 This many moles of Carbon would be necessary to produce 38 moles of Carbon Monoxide from the equation: SnO2 + 2C  Sn + 2CO

  10. CATEGORY 1 - $300 This many moles of Copper Nitrate would be produced from 28 moles of HNO3 from the equation: Cu + 4HNO3 Cu(NO3)2 + 2 NO2 + 2H2O

  11. CATEGORY 1 - $400 This many grams of Carbon would be necessary to produce 8 moles of Carbon Dioxide in the equation: 2C + FeCr2O4 FeCr2 + 2CO2

  12. CATEGORY 1 - $500 This many moles of Iron Oxide would be produced from 446.8 g of Iron from the equation: 4Fe + 3O2 2Fe2O3

  13. CATEGORY 2 - $100 This many grams of CH4 is needed to produce 50 grams of CHCl3 from the reaction: CH4 + 3Cl2 CHCl3 + 3 HCl

  14. CATEGORY 2 - $200 This many grams of Carbon Dioxide are produced from 100 grams of C2H5OH from the equation: C2H5OH + 3O2 2CO2 + 3H2O

  15. CATEGORY 2 - $300 This many grams of Lead Sulfate is produced from 25 grams of lead in the following equation: Pb + PbO2 2H2SO4 + 2PbSO4 + 2H2O

  16. CATEGORY 2 - $400 This many grams of Hydrogen are needed to produce 45 grams of CH3OH from the equation: CO + 2H2 CH3OH

  17. CATEGORY 2 - $500 This many grams of Gold is able to be extracted from 25 grams of NaCN from the equation: 4Au + 8NaCN + O2 + H2O 4NaAu(CN)2 + 4NaOH

  18. CATEGORY 3 - $100 This is the limiting reactant for the production of Zn(OH)2 between if 25 grams of Zinc and 30 grams of MnO2 in this equation: Zn + 2MnO2 + H2O  Zn(OH)2 + Mn2O3

  19. CATEGORY 3 - $200 This is the limiting reactant for the production of PCl5 between 23 grams of P4 and 16 g Cl2 for the equation: P4 + 10Cl2 4PCl5

  20. CATEGORY 3 - $300 This is the limiting reactant between 25 grams of Lithium and 25 grams of Bromine Liquid to produce LiBr. 2 Li + Br2 2 LiBr

  21. CATEGORY 3 - $400 This many moles of Iron (II) Hydroxide are produced if 5 moles of Iron and 8 moles NiO(OH) react, from this equation: Fe + 2NiO(OH) +2H2O  Fe(OH)2 + 2Ni(OH)2

  22. CATEGORY 3 - $500 This many moles of Cesium Xenon Heptafluoride (CsXeF7), can be produced from the reaction of 12.5 moles of Cesium Fluoride (CsF) with 10 moles of Xenon Hexafluoride (XeF6): CsF + XeF6 CsXeF7

  23. CATEGORY 4 - $100 This is the empirical formula for a compound containing 56% Vanadium and 44% Oxygen.

  24. CATEGORY 4 - $200 This is the empirical for a compound containing 71.65% Chlorine, 24.27% Oxygen and 4.07% Hydrogen.

  25. CATEGORY 4 - $300 This is how many electrons the Chloride anion would have.

  26. CATEGORY 4 - $400 This is the coefficient that be needed in front of O2, in order to balance the equation below. ___C6H14 + ___O2 ___CO2 + ___H2O

  27. CATEGORY 4 - $500 This is the empirical formula of a compound containing 38.67% Carbon, 16.22% Hydrogen and the rest is Nitrogen.

  28. CATEGORY 5 - $100 The artist or the name of this song.

  29. CATEGORY 5 - $200 The artist or the name of this song.

  30. CATEGORY 5 - $300 The artist or the name of this song.

  31. CATEGORY 5 - $400 The artist or the name of this song.

  32. CATEGORY 5 - $500 The artist or the name of this song.

  33. CATEGORY 1 - $100 WHAT IS 6 moles ?

  34. CATEGORY 1 - $200 WHAT IS 38 moles ?

  35. CATEGORY 1 - $300 WHAT IS 7 moles ?

  36. CATEGORY 1 - $400 WHAT IS 96 grams ?

  37. CATEGORY 1 - $500 WHAT IS 4 moles ?

  38. CATEGORY 2 - $100 WHAT IS 6.72 grams ?

  39. CATEGORY 2 - $200 WHAT IS 191 grams ?

  40. CATEGORY 2 - $300 WHAT IS 73.2 grams ?

  41. CATEGORY 2 - $400 WHAT IS 5.67 grams ?

  42. CATEGORY 2 - $500 WHAT IS 50.2 grams ?

  43. CATEGORY 3 - $100 WHAT IS MnO2 ?

  44. CATEGORY 3 - $200 WHAT IS Cl2 ?

  45. CATEGORY 3 - $300 WHAT IS Br2 ?

  46. CATEGORY 3 - $400 WHAT IS 4 moles ?

  47. CATEGORY 3 - $500 WHAT IS 10 moles ?

  48. CATEGORY 4 - $100 WHAT IS V2O5 ?

  49. CATEGORY 4 - $200 WHAT IS Cl4O3H8 ?

More Related