1. Stoichiometry – Ch. 12.3

2. What would be produced if two pieces of bread and a slice of salami reacted together? + ?

4. 2B + S

15. +

16. + 2B + S B 2 S

17. + 2B + 1S 1B 2 S

18. + Bread (B) and Salami (S) react in a 2 to 1 (2:1) ratio to produce sandwich (B2S)

19. + 6B + 2S ?

20. +

30. + + (leftover)

31. 6B + 2S + 2B 2 S

32. 6B + 2S + 2B 2 S The amount of salami limits the amount of sandwiches that can be produced

33. WHY?

34.  Available Ingredients  4 slices of bread  1 jar of peanut butter  1/2 jar of jelly b Limiting Reactant bread b Excess Reactants peanut butter and jelly

35.  Limiting Reactant  used up in a reaction  determines the amount of product  Excess Reactant  Left over in a reaction  added to ensure that the other reactant is completely used up

54. 2H 2 + 1 O 2 2H 2 O

55. ? what happens if there is an excess amount of hydrogen?

71. ?

72. ?

73. ? 5H 2 + 1 O 2 2H 2 O+ 3H 2

74. 1. Write a balanced equation. 2. For each reactant, calculate mol present and find ratio. 3. Smaller answer indicates:  limiting reactant  amount of product 4. Use limiting reactant to find mol of product and needed reactant 5. Convert mol to grams

75.  Determining limiting reactant  Calculating product  Determining excess reactant

76.  Determine the mass of tetraphosphorous decoxide formed if 25.0 g of phosphorus (P 4 ) and 50.0 g of oxygen are combined?  First, write the equation… P 4 + O 2  P 4 O 10  Then balance it… P 4 + 5O 2  P 4 O 10

77.  Determine moles of each reactant 25.0 g P 4 x 1 mol = 0.202 mol P 4 123.9 g 50.0 g O 2 x 1 mol = 1.56 mol O 2 32.0 g 25.0 g50.0 g

78.  Calculate mole ratio 1.56 mol O 2 = 7.72 mol O 2 0.202 mol P 4 1 mol P 4  Determine mol ratio from equation 5 mol O 2 1 mol P 4

79.  Because 7.72 mol O 2 is available but only 5 mol is needed to react with 1 mol P 4, O 2 is in excess and P 4 is the limiting reactant. Use mol of P 4 to determine mol of P 4 O 10 produced.  Answer #1: P is limiting and O is excess  Multiply limiting reactant (P 4 ) by mole ratio 0.202 mol P 4 x 1 mol P 4 O 10 = 0.202 mol P 4 O 10 1 mol P 4

80.  Convert mol P 4 O 10 to grams 0.202 mol P 4 O 10 x 283.9 g P 4 O 10 = 57.3 g P 4 O 10 1 mol P 4 O 10  Answer #2: 57.3 g P 4 O 10 is produced

81.  O 2 is in excess so only so much is used.  Use P 4 to determine mol and mass of O 4 used 0.202 mol P 4 x 5 mol O 2 = 1.01 mol O 2 needed 1 mol P 4  Multiply mol O 2 by molar mass 1.01 mol O 2 x 32.0 g O 2 = 32.3 g O 2 needed 1 mol O 2

82.  Subtract the mass of O 2 needed from the mass available to calculate excess O 2. 50.0 g O 2 available – 32.3 g needed = 17.7 g O 2 Answer #3: 17.7 g O 2 is left over (in excess)

83. 6Na + Fe 2 O 3  3Na 2 0 + 2Fe  Find:  The limiting reactant  The reactant in excess  The mass of solid iron produced  The mass of excess reactant that remains

84. calculated on paper measured in lab

85.  When 45.8 g of K 2 CO 3 react with excess HCl, 46.3 g of KCl are formed. Calculate the theoretical and % yields of KCl. K 2 CO 3 + 2HCl  2KCl + H 2 O + CO 2 45.8 g? g actual: 46.3 g

86. Theoretical Yield: 45.8 g K 2 CO 3 1 mol K 2 CO 3 138.21 g K 2 CO 3 = 49.4 g KCl 2 mol KCl 1 mol K 2 CO 3 74.55 g KCl 1 mol KCl K 2 CO 3 + 2HCl  2KCl + H 2 O + CO 2 45.8 g? g actual: 46.3 g

87. Theoretical Yield = 49.4 g KCl % Yield = 46.3 g 49.4 g  100 = 93.7% K 2 CO 3 + 2HCl  2KCl + H 2 O + CO 2 45.8 g49.4 g actual: 46.3 g