1. Stoichiometry! is the calculation of relative quantities of reactants and products in chemical reactions. SOURCE: Hands on Chemistry Activities…Herr Stoichiometry is found in every branch of science!

2. Starting with Water! 2H 2 + O 2  2H 2 0 Have No Fear Of Ice Cold Beer LEFTRIGHT HOHO 2221 2242 4242 You have to balance equations by changing Coefficients! H H O O H H H H O O H H + O O H H O O H H H H H H

3. Reading Chemical Equations! 2H 2 + O 2  2H 2 0 What does this mean? Two molecules of H 2 gas combine with One molecule of O 2 gas to produce Two molecules of Water.

4. Reading Chemical Equations! 2H 2 + O 2  2H 2 0 What does this mean? Two molecules of H 2 gas combine with One molecule of O 2 gas to produce Two molecules of Water. 2H 2 + O 2  2H 2 0 Two MOLES of H 2 gas combine with One MOLE of O 2 gas to produce Two MOLES of Water.

5. Reading Chemical Equations! 2H 2 + O 2  2H 2 0 What does this mean? Two molecules of H 2 gas combine with One molecule of O 2 gas to produce Two molecules of Water. 2H 2 + O 2  2H 2 0 Two MOLES of H 2 gas combine with One MOLE of O 2 gas to produce Two MOLES of Water. Four GRAMS of H 2 gas combine with Thirty-Two GRAMS of O 2 gas to produce Thirty-Six GRAMS of Water.

6. Reading Chemical Equations! CH 4 + 2O 2  CO 2 + 2H 2 0 What does this mean?

7. Reading Chemical Equations! CH 4 + 2O 2  CO 2 + 2H 2 0 What does this mean? One molecule of CH 4 (methane) combines with Two molecules of O 2 gas to produce One molecule of CO 2 (carbon dioxide gas) and Two molecules of Water. SOURCE Wiki: Stoichiometry

8. Reading Chemical Equations! CH 4 + 2O 2  CO 2 + 2H 2 0 What does this mean? One molecule of CH 4 (methane) combines with Two molecules of O 2 gas to produce One molecule of CO 2 (carbon dioxide gas) and Two molecules of Water. SOURCE Wiki: Stoichiometry One MOLE of CH 4 (methane) combines with Two MOLES of O 2 gas to produce One MOLE of CO 2 (carbon dioxide gas) and Two MOLES of Water.

9. Reading Chemical Equations! CH 4 + 2O 2  CO 2 + 2H 2 0 What does this mean? One molecule of CH 4 (methane) combines with Two molecules of O 2 gas to produce One molecule of CO 2 (carbon dioxide gas) and Two molecules of Water. SOURCE Wiki: Stoichiometry One MOLE of CH 4 (methane) combines with Two MOLES of O 2 gas to produce One MOLE of CO 2 (carbon dioxide gas) and Two MOLES of Water. Sixteen GRAMS of CH 4 combines with Sixty-Four GRAMS of O 2 gas to produce Fourty-Four GRAMS of CO 2 and Thirty-Six GRAMS of Water.

10. x = Unit you want to change Unit you are changing into Conversion Factor How many feet is 744 inches?

11. x = Unit you want to change Unit you are changing into Conversion Factor How many moles is 400g of Sodium?

12. UNIT CANCELLING WORKSHEET MOLE RATIO WORKSHEET

13. WATER: Stoichiometry 2H 2 + O 2  2H 2 0 If you have 80 molecules of H 2, how much O 2 do you need to react with all of it and how much water would you get?

14. WATER: Stoichiometry 2H 2 + O 2  2H 2 0 If you have 80 molecules of H 2, how much O 2 do you need to react with all of it and how much water would you get?

15. WATER: Stoichiometry 2H 2 + O 2  2H 2 0 If you have 80 molecules of H 2, how much O 2 do you need to react with all of it and how much water would you get?

16. WATER: Stoichiometry 2H 2 + O 2  2H 2 0 If you have 80 molecules of H 2, how much O 2 do you need to react with all of it and how much water would you get?

17. WATER: Stoichiometry 2H 2 + O 2  2H 2 0 If you have 80 MOLES of H 2, how much O 2 do you need to react with all of it and how much water would you get?

18. RUST: Iron (?) Oxide 2Fe + 3O 2  2Fe 2 0 3 If you have 444g of Iron, how many moles of O 2 will you need to react with all of it and how much rust will you end up with? First Convert Grams into Moles!!!

19. RUST: Iron (?) Oxide 2Fe + 3O 2  2Fe 2 0 3 If you have 444g of Iron, how many moles of O 2 will you need to react with all of it and how much rust will you end up with? 26 Fe IRON 55.85 First Convert Grams into Moles!!!

20. RUST: Iron (?) Oxide 2Fe + 3O 2  2Fe 2 0 3 If you have 444g of Iron, how many moles of O 2 will you need to react with all of it and how much rust will you end up with? 26 Fe IRON 55.85 First Convert Grams into Moles!!! Molar Mass = 55.85g/mol

21. RUST: Iron (?) Oxide 4Fe + 3O 2  2Fe 2 0 3 If you have 444g of Iron, how many moles of O 2 will you need to react with all of it and how much rust will you end up with? 26 Fe IRON 55.85 First Convert Grams into Moles!!! Molar Mass = 55.85g/mol

22. RUST: Iron (?) Oxide 4Fe + 3O 2  2Fe 2 0 3 If you have 444g of Iron, how many moles of O 2 will you need to react with all of it and how much rust will you end up with? 26 Fe IRON 55.85 Molar Mass = 55.85g/mol

23. RUST: Iron (?) Oxide 4Fe + 3O 2  2Fe 2 0 3 If you have 444g of Iron, how many moles of O 2 will you need to react with all of it and how much rust will you end up with? 26 Fe IRON 55.85 Molar Mass = 55.85g/mol

24. RUST: Iron (?) Oxide 4Fe + 3O 2  2Fe 2 0 3 If you have 444g of Iron, how many moles of O 2 will you need to react with all of it and how much rust will you end up with? 26 Fe IRON 55.85 Molar Mass = 55.85g/mol

25. RUST: Iron (?) Oxide 4Fe + 3O 2  2Fe 2 0 3 If you have 444g of Iron, how many moles of O 2 will you need to react with all of it and how much rust will you end up with? 26 Fe IRON 55.85

26. RUST: Iron (?) Oxide 4Fe + 3O 2  2Fe 2 0 3 If you have 444g of Iron, how many moles of O 2 will you need to react with all of it and how much rust will you end up with? 26 Fe IRON 55.85

27. Photochemical Smog!

29. Assignment: 1.Hydrogen Sulfide which smells like rotten eggs is found in volcanic ash. The balanced equation for the burning of hydrogen sulfide is 2H 2 S(g) + 3O 2 (g)  2SO 2 (g) + 2H 2 O(g). Express this equation using the masses of all the reactants and products. 2.If you were given 7.5 x 10 24 atoms of aluminum, how many moles of oxygen would you need so that all of the aluminum forms aluminum oxide? How many moles of aluminum oxide would be produced? ( Hint: atoms to moles via avogadro’s number and write a balanced equation for the formation of aluminum oxide). 3.How many molecules of oxygen are produced when 29.2 grams of H 2 O is decomposed by electrolysis into diatomic hydrogen and oxygen gas? 4.How many moles of ammonia are produced when 0.6mol nitrogen reacts with hydrogen? N 2 + 3H 2  2NH 3 5.Calculate the number of grams of NH 3 produced when 5.4grams of hydrogen reacts with an excess of nitrogen. 6.Phosphorous and hydrogen can be combined to form phosphine (PH 3 ) according to the following equation: P 4 (s) + 6H 2 (g)  4PH 3 (g). How many liters of phosphine are formed when 0.42L of hydrogen reacts with an excess of phosphorous?

31. A Chemical Formula is Like a Recipe: BROWNIES 2 c W + 4 c S + 4 c F + 4E + 8 oz B + 8 pc C  10 Brownies

32. A Chemical Formula is Like a Recipe: BROWNIES We know that 4 eggs combined with everything else yields 10brownies but what if you only had 3 eggs? 2 c W + 4 c S + 4 c F + 4E + 8 oz B + 8 pc C  10 Brownies

33. A Chemical Formula is Like a Recipe: BROWNIES We know that 4 eggs combined with everything else yields 10brownies but what if you only had 3 eggs? You have ¾ of the needed eggs so cut the entire recipe by ¾ and end up with ¾ the brownies. 2 c W + 4 c S + 4 c F + 4E + 8 oz B + 8 pc C  10 Brownies

34. A Chemical Formula is Like a Recipe: BROWNIES 2 c W + 4 c S + 4 c F + 4E + 8 oz B + 8 pc C  10 Brownies

35. How many Brownies can you make? What limits your from making more?

36. A Chemical Formula is Like a Recipe: BROWNIES Sometimes in chemistry we don’t have all the necessary ingredients for complete reactions! 2 c W + 4 c S + 4 c F + 4E + 8 oz B + 8 pc C  10 Brownies

37. Water 2H 2 + O 2  2H 2 0 Suppose you have 50moles of Oxygen and 40 moles of hydrogen. How much water can you make?

38. Water 2H 2 + O 2  2H 2 0 Suppose you have 50moles of Oxygen and 40 moles of hydrogen. How much water can you make? By looking at the balanced equation we know that you need 2 moles of hydrogen for every mole of oxygen but we only have 40 moles of hydrogen.

39. Water 2H 2 + O 2  2H 2 0 Suppose you have 50moles of Oxygen and 40 moles of hydrogen. How much water can you make? By looking at the balanced equation we know that you need 2 moles of hydrogen for every mole of oxygen but we only have 40 moles of hydrogen.

40. Water 2H 2 + O 2  2H 2 0 Suppose you have 50moles of Oxygen and 40 moles of hydrogen. How much water can you make? By looking at the balanced equation we know that you need 2 moles of hydrogen for every mole of oxygen but we only have 40 moles of hydrogen. Only 20 moles of the Oxygen will react and there will be an excess of 30 moles leftover.

41. Water 2H 2 + O 2  2H 2 0 Suppose you have 50moles of Oxygen and 40 moles of hydrogen. How much water can you make? By looking at the balanced equation we know that you need 2 moles of hydrogen for every mole of oxygen but we only have 40 moles of hydrogen. Only 20 moles of the Oxygen will react and there will be an excess of 30 moles leftover. Hydrogen: limiting reagent Oxygen: excess reagent

42. Limiting and Excess limiting reagent is the reagent that determines the amount of product that can be formed by a reaction The reagent that is not used up is called the excess reagent.

43. Copper Reacts with sulfur to form copper (I) sulfide according to the following balanced equation. 2Cu(s) + S(s)  Cu 2 S(s) What is the limiting reagent when 80g Cu reacts with 25g S?

44. You need twice as much Copper as sulfur in terms of moles but our problem gives us values in grams. First we need to convert 80g Cu  moles and 25g S  moles. You need twice as much Copper as sulfur in terms of moles but our problem gives us values in grams. First we need to convert 80g Cu  moles and 25g S  moles.

45. Copper Reacts with sulfur to form copper (I) sulfide according to the following balanced equation. 2Cu(s) + S(s)  Cu 2 S(s) What is the limiting reagent when 80g Cu reacts with 25g S?

47. We Know that we need twice as much copper as sulfur from the balanced equation. We don’t have twice as much so Copper is the limiting reagent! There will be an excess of sulfur.

48. Copper Reacts with sulfur to form copper (I) sulfide according to the following balanced equation. 2Cu(s) + S(s)  Cu 2 S(s) What is the maximum amount of Cu 2 S that can be formed in grams when 80g of Cu reacts with 25g of S?

49. We know that Copper is the limiting reagent and there will be an excess of sulfur. So ALL of the copper will react!

50. Copper Reacts with sulfur to form copper (I) sulfide according to the following balanced equation. 2Cu(s) + S(s)  Cu 2 S(s) What is the maximum amount of Cu 2 S that can be formed in grams when 80g of Cu reacts with 25g of S? We know that Copper is the limiting reagent and there will be an excess of sulfur. So ALL of the copper will react!

51. Copper Reacts with sulfur to form copper (I) sulfide according to the following balanced equation. 2Cu(s) + S(s)  Cu 2 S(s) What is the maximum amount of Cu 2 S that can be formed in grams when 80g of Cu reacts with 25g of S? We know that Copper is the limiting reagent and there will be an excess of sulfur. So ALL of the copper will react! Convert Moles  Grams!

52. Copper Reacts with sulfur to form copper (I) sulfide according to the following balanced equation. 2Cu(s) + S(s)  Cu 2 S(s) What is the maximum amount of Cu 2 S that can be formed in grams when 80g of Cu reacts with 25g of S? We know that Copper is the limiting reagent and there will be an excess of sulfur. So ALL of the copper will react! Convert Moles  Grams!

53. Percent Yield The theoretical yield is the maximum amount of product that could be formed from given amounts of reactants. In contrast, the amount of product that actually forms when the reaction is carried out in the laboratory is called the actual yield.

58. Assignment: