# 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 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:

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: 562-985-5350

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

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 2 32.0 g O 2 36.0 g H 2 O

Stoichiometric Relationships

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

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

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

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

Limiting Reactant

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

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

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)

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

Excess Reactant We can also calculate the amount of the non- limiting reactant that is used up, and thus how much is left over 92.72 g Cl 2 are required to react with all the P 4 Thus 95.0-92.72=2.3 g Cl 2 are left over as excess

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

Actual Yield The amount of product actually produced in a reaction

Percent Yield actual yield % yield =  100% theoretical yield For example, if only 103.5 g of PCl 3 were actually produced

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 = 0.451 kg N 2 H 4

EXAMPLE (b) What is the actual yield if 0.299 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) = 0.293 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 = 0.451 kg N 2 H 4 (b) actual yield

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= 0.451 kg N 2 H 4 (b) actual yield= 0.293 kg N 2 H 4 (c) percent yield 0.293 kg % yield =  100% = 65.0 % yield 0.451kg

1.16 % 2.32 % 3.50 % 4.65 %

Combustion Analysis

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 = = 0.318-mg H (18.01-mg H 2 O )

# mg O = 6.49 - 4.48 - 0.318 = 1.69 mg O Empirical Formula C7H6O2C7H6O2

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