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Free Group Study Tutoring Monday 5-6pm Tuesday 5-6pm Friday 11am-12pm Tutoring will begin Friday Sept. 16 th at 11am in the Learning Assistance Center!

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Presentation on theme: "Free Group Study Tutoring Monday 5-6pm Tuesday 5-6pm Friday 11am-12pm Tutoring will begin Friday Sept. 16 th at 11am in the Learning Assistance Center!"— Presentation transcript:

1 Free Group Study Tutoring Monday 5-6pm Tuesday 5-6pm Friday 11am-12pm Tutoring will begin Friday Sept. 16 th at 11am in the Learning Assistance Center! For more information call the LAC at:

2 Stoichiometry stoi·chi·om·e·try noun Literally – measuring the components Calculations of the quantitative relationships between reactants and products in a chemical reaction.

3 The Mole and Chemical Reactions: The Macro-Nano Connection 2 H 2 + O 2  2 H 2 O 2 H 2 molecules 1 O 2 molecule 2 H 2 O molecules 2 moles H 2 molecules 1 mole O 2 molecules 2 moles H 2 O molecules 4.0 g H g O g H 2 O

4 Stoichiometric Relationships

5 EXAMPLE How much iron(III) oxide and aluminum powder are required to field weld the ends of two rails together? Assume that the rail is 132 lb Fe/yard and that the weld is 1/10 inch wide. weld Photo by Mike Condren

6 The mass of iron in 1/10 inch of this rail is: Fe 2 O Al  Al 2 O 3 +2 Fe

7 1.C 2.Ge 3.Si 4.Sn 5.Ti

8 Limiting Reactant reactant that limits the amount of product that can be produced

9 Limiting Reactant

10 EXAMPLE What is the number of moles of CaSO 4 (S) that can be produced by allowing 1.0 mol SO 2, 2.0 mol CaCO 3, and 3.0 mol O 2 to react? 2SO 2(g) + 2CaCO 3(s) + O 2(g)  2CaSO 4(S) + 2CO 2(g) balanced equation relates: 2SO 2(g)  2CaCO 3(s)  O 2(g) have only: 1SO 2(g)  2CaCO 3(s)  3O 2(g) not enough SO 2 to use all of the CaCO 3 or the O 2 not enough CaCO 3 to use all of the O 2 SO 2 is the limiting reactant

11 Compare Amounts of Product SO 2 limits the amount of product formed

12 Mass of Product 95.0 g of chlorine and 27.0 g of phosphorus react to form PCl 3. What mass of PCl 3 is formed? P 4 (s) + 6 Cl 2 (g)  4 PCl 3 (l)

13 27.0g phosphorus gives the smaller yield. Phosphorus is the limiting reactant it limits the amount of product formed

14 Excess Reactant We can also calculate the amount of the non- limiting reactant that is used up, and thus how much is left over g Cl 2 are required to react with all the P 4 Thus =2.3 g Cl 2 are left over as excess

15 Theoretical Yield The amount of product produced by a reaction based on the amount of the limiting reactant

16 Actual Yield The amount of product actually produced in a reaction

17 Percent Yield actual yield % yield =  100% theoretical yield For example, if only g of PCl 3 were actually produced

18 EXAMPLE A rocket fuel, hydrazine, is produced by a reaction of Cl 2 with excess NaOH and NH 3. (a) What theoretical yield can be produced from 1.00 kg of Cl 2 ? 2NaOH + Cl 2 + 2NH 3  N 2 H 4 + 2NaCl + 2H 2 O (a) to calculate the theoretical yield, use the net equation for the overall process (1.00 kg Cl 2 ) (1 kmol Cl 2 ) (70.9 kg Cl 2 ) molar mass (1 kmol N 2 H 4 ) (1 kmol Cl 2 ) balanced equation (32.0 kg N 2 H 4 ) (1 kmol N 2 H 4 ) molar mass = kg N 2 H 4

19 EXAMPLE (b) What is the actual yield if kg of 98.0% N 2 H 4 is produced for every 1.00 kg of Cl 2 ? (98.0 kg N 2 H 4 ) (100 kg product) purity factor (0.299 kg product) = kg N 2 H 4 2NaOH + Cl 2 + 2NH 3  N 2 H 4 + 2NaCl + 2H 2 O (a) theoretical yield#kg N 2 H 4 = kg N 2 H 4 (b) actual yield

20 EXAMPLE (c) What is the percent yield of pure N 2 H 4 ? 2NaOH + Cl 2 + 2NH 3  N 2 H 4 + 2NaCl + 2H 2 O (a) theoretical yield= kg N 2 H 4 (b) actual yield= kg N 2 H 4 (c) percent yield kg % yield =  100% = 65.0 % yield 0.451kg

21 1.16 % 2.32 % 3.50 % 4.65 %

22 Combustion Analysis

23 Example Benzoic acid is known to contain only C, H, and O. A 6.49-mg sample of benzoic acid was burned completely in a C-H analyzer. The increase in the mass of each absorption tube showed that 16.4-mg of CO 2 and 2.85-mg of H 2 O formed. What is the empirical formula of benzoic acid? (16.4-mg of CO 2 )(12.01-mg C) # mg C = = 4.48-mg C (44.01-mg CO 2 ) (2.85-mg of H 2 O )(2.01-mg H) # mg H = = mg H (18.01-mg H 2 O )

24 # mg O = = 1.69 mg O Empirical Formula C7H6O2C7H6O2


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