Chemistry Lecture 93 B. Rife CHS

page 6/6

Chemistry Lecture ’93 B. Rife CHS

Text: Modern Chemistry; Holt, Rinehart & Winston 1993

Stoichiometry Chapter 9

Homework: DUE DATE

1 Section Reviews (pg 265,271,278)

2 Reviewing Concepts: (1 - 11) (pg 279)

3 Problems (all) (pg 279-281)

4 Chapter/Section Review (Handout)

Exam Date _

9.1 Introduction to Stoichiometry

9.1A Define stoichiometry and distinguish between composition and reaction stoichiometry ( )

STOICHIOMETRY - DESCRIPTION OF THE QUANTITATIVE RELATIONSHIPS AMONG ELEMENTS IN COMPOUNDS (COMPOSITION STOICHIOMETRY), AND AMONG SUBSTANCES AS THEY UNDERGO CHEMICAL CHANGES. (REACTION STOICHIOMETRY)

- THE BRANCH OF CHEMISTRY THAT DEALS WITH THE NUMERICAL RELATIONSHIPS OF ELEMENTS AND COMPOUNDS AND THE MATHEMATICAL PROPORTIONS OF REACTANTS AND PRODUCTS IN CHEMICAL REACTIONS.

9.1B define mole ratio and describe its role in stoichiometry calculations. ( )

THE COEFFICIENTS OF A BALANCED EQUATION REPRESENT NUMBERS OF MOLES OF REACTANTS AND PRODUCTS.

MOLE RATIO - IS THE RATIO OF MOLES OF ONE SUBSTANCE TO MOLES OF ANOTHER SUBSTANCE IN A BALANCED CHEMICAL EQUATION

9.1C Give the mole ratio for any two substances in a chemical equation ( )

THE STEPS FOR CALCULATING MASSES OF REACTANTS AND PRODUCTS IN CHEMICAL REACTIONS (MASS - MASS)

1. BALANCE THE EQUATION FOR THE REACTION

2. CONVERT THE KNOWN MASS OF THE REACTANT OR PRODUCT TO MOLES OF THAT SUBSTANCE.

(CONVERSION FACTOR IS THE MOLAR MASS)

3. USE THE BALANCED EQUATION TO SET UP THE APPROPRIATE MOLE RATIO (CONVERSION FACTOR)

4. USE THE APPROPRIATE MOLE RATIOS TO CALCULATE THE NUMBER OF MOLES OF THE DESIRED REACTANT OR PRODUCT.

5. CONVERT FROM MOLES BACK TO GRAMS IF REQUIRED BY THE PROBLEM. (CONVERSION FACTOR IS THE MOLAR MASS)

EXAMPLE: COMBUSTION OF PROPANE GAS

C3H8 (g) + 5O2 (g) ---> 3CO2 (g) + 4H2O (g)

MOLE RATIOS: 5 mol O2 3 mol CO2

1 mol C3H8 1 mol C3H8

9.1D Name the four types of reaction-stiometry calculations ( )

GIVEN ASKED FOR CONVERSION FACTORS

MOLE - MOLE MOLE RATIO

MOLE - MASS MOLE RATIO MOLAR MASS

MASS - MOLE MOLAR MASS MOLE RATIO

MASS - MASS MOLAR MASS MOLE RATIO MOLAR MASS

9.2 Ideal Stoichiometric Calculations

IDEAL CONDITIONS IMPLIES THE COMPLETE CONVERSION OF ALL REACTANTS INTO PRODUCTS.

STOICHIOMETRIC QUANTITIES REFER TO RELATIVE AMOUNTS OF REACTANTS THAT ARE IN THE SAME MOLE RATIO AS IMPLIED BY THE BALANCED EQUATION FOR A CHEMICAL REACTION.

MOLE-MOLE CALCULATIONS

9.2A Calculate the amount in moles of a reactant or product, given the amount in moles of a different reactant or product ( )

GIVEN DATA x CONVERSION FACTOR(S) ---> REQUIRED DATA

MOLES A x MOLES B = MOLES B

MOLES A

EXAMPLE: COMBUSTION OF PROPANE GAS

C3H8 (g) + 5O2 (g) ---> 3CO2 (g) + 4H2O (g)

HOW MANY MOLES OF OXYGEN IS REQUIRED TO COMBUST 3 MOLES OF PROPANE?

3 MOLES C3H8 x 5 mol O2 = 15 mol O2

1 mol C3H8

MOLE - MASS CALCULATIONS

9.2B Calculate the mass of a reactant or product, given the amount in moles of a different reactant or product ( )

MOLES A x MOLES B x MOLAR MASS B = MASS B

MOLES A MOLES B

EXAMPLE: COMBUSTION OF PROPANE GAS

C3H8 (g) + 5O2 (g) ---> 3CO2 (g) + 4H2O (g)

WHAT MASS OF WATER IS PRODUCED IN THE COMBUSTION OF 5 MOLES OF PROPANE?

5 MOLES C3H8 x 4 mol H2O x 18 g H2O = 360 g H2O

1 mol C3H8 1 mol H2O

MASS - MOLE CALCULATIONS

MASS A x MOLES A x MOLES B = MOLES B

MOLAR MASS A MOLES A

EXAMPLE: COMBUSTION OF PROPANE GAS

C3H8 (g) + 5O2 (g) ---> 3CO2 (g) + 4H2O (g)

HOW MANY MOLES OF OXYGEN IS REQUIRED IN THE COMBUSTION OF PROPANE TO PRODUCE 180 g of CO2?

180 g CO2 x 1 mol CO2 x 5 mol O2 = 6.8 mol O2

44 g CO2 3 mol CO2

MASS - MASS

9.2D Calculate the mass of a reactant or product, given the mass of a different reactant or product. ( )

MASS A x MOLES A x MOLES B x MOLAR MASS B = MASS B

MOLAR MASS A MOLES A MOLES B

EXAMPLE #1: COMBUSTION OF OCTANE GAS

2 C8H18 (l) + 25 O2 (g) ---> 16 CO2 (g) + 18 H2O (g)

HOW MANY KILOGRAMS OF CARBON DIOXIDE IS PRODUCED IN THE COMBUSTION OF ONE LITER (KILOGRAM) OF OCTANE?

1 kg C8H18 x 1000 g C3H8 = 1000 g C3H8

1 kg C3H8

1000 g C3H8 x 1 mol C8H18 x 16 mol CO2 x 44 g CO2 = 3087.7 g CO2

114 g C8H18 2 mol C3H8 1 mol CO2

3087.8 g CO2 x 1 kg CO2 = 3.1 kg CO2

1000 g CO2

EXAMPLE #2: A MASS OF 25.0 g OF MERCURY(II) OXIDE IS DECOMPOSED BY HEATING (JOSEPH PRIESTLEY).

THE REACTION IS 2HgO (s) _ 2Hg (l) + O2 (g)

A. HOW MANY MOLES OF MERCURY(II) OXIDE ARE DECOMPOSED?

25g HgO ( 1 MOL HgO ) = 0.115 MOL MERCURY(II) OXIDE

216.6g HgO

B. HOW MANY MOLES OF OXYGEN ARE PREPARED?

25g HgO ( 1 MOL HgO ) ( 1 mol O2 ) = 0.0575 MOL OXYGEN

216.6g HgO 2 mol HgO

C. HOW MANY GRAMS OF OXYGEN ARE PREPARED?

25g HgO ( 1 MOL HgO ) ( 1 mol O2 ) ( 32g O2 ) = 1.84 g O2

216.6g HgO 2 mol HgO 1 mol O2

9.3 Limiting Reactants and Percent Yield

9.3A Define limiting reactant ( )

THE LIMITING REACTANT IN A REACTION IS THE REACTANT THAT IS CONSUMED COMPLETELY. THE QUANTITY OF OTHER REACTANT(S) CONSUMED AND PRODUCT(S) FORMED DEPENDS ON THE QUANTITY OF THE LIMITING REAGENT.

THE SUBSTANCE THAT IS NOT USED UP COMPLETELY IN A REACTION IS SOMETIMES CALLED THE EXCESS REACTANT.

9.3B Describe the method for determining which of two reactants is a limiting reactant. ( )

TO CALCULATE THE LIMITING REAGENT:

1. CALCULATE THE AMOUNT OF PRODUCT THAT WOULD BE FORMED IF EACH REACTANT WAS COMPLETELY CONSUMED.

2. CHOOSE THE SMALLEST OF PRODUCT AMOUNTS CALCULATED. THE REACTANT THAT PRODUCES THE SMALLEST AMOUNT IS THE LIMITING REACTANT.

TEXT METHOD

1. CHOOSE REACTANT A

2. CALCULATE THE NUMBER OF MOLES OF REACTANT B, THAT IS REQUIRED IF REACTANT A IS USED UP.

3. IF MORE IS REQUIRED (CALCULATED) THAN IS INDICATED BY THE EQUATION, THEN REACTANT B IS THE LIMITING REACTANT.

IF LESS IS REQUIRED (CALCULATED) THAN IS INDICATED BY THE EQUATION, THEN REACTANT A IS THE LIMITING REACTANT.

9.3C Calculate the amount in moles or mass of a product, given the amounts in moles or masses of two reactants, one of which is in excess. ( )

EXAMPLE

PHOSPHORUS TRICHLORIDE, PCl3, IS A COMMERCIALLY IMPORTANT COMPOUND IN THE MANUFACTURE OF PESTICIDES AND GASOLINE. IT IS MADE BY THE DIRECT COMBINATION OF PHOSPHORUS AND CHLORINE.

P4 (l) + 6 Cl2 (g) ---> 4 PCl3 (l)

WHAT MASS OF PCl3 (l) FORMS IN THE REACTION OF 125 g P4 WITH 325 g Cl2 ?

SOLUTION:

125 g P4 x 1 mol P4 x 4 mol PCl3 = 4.04 mol PCl3

123.9 g P4 1 mol P4

325 g Cl2 x 1 mol Cl2 x 4 mol PCl3 = 3.06 mol PCl3

70.91 g Cl2 6 mol Cl2

3.06 mol IS LESS THAN 4.04 mol PCl3 THUS Cl2 IS THE LIMITING REACTANT

3.06 mol PCl3 137.3 g PCl3 = 420 g PCl3 FORMED

1 mol PCl3

TEXT SOLUTION:

1. P4 IS REACTANT A

2. 125 g P4 x 1 mol P4 x 6 mol Cl2 = 6.05 mol Cl2

123.9 g P4 1 mol P4

3. IF MORE IS REQUIRED ( 6.05 CALCULATED) THAN IS INDICATED BY THE EQUATION ( 6 ), THEN REACTANT B (Cl2) IS THE LIMITING REACTANT.

9.3D Define theoretical yield, actual yield, and percent yield. ( )

THEORETICAL YIELD - ASSUMES THE REACTION GOES TO COMPLETION AND NO PRODUCT IS LOST.

- IS THE MAXIMUM AMOUNT OF PRODUCT THAT CAN BE PRODUCED FROM A GIVEN AMOUNT OF REACTANTS.

ACTUAL YIELD - IS THE MEASURED AMOUNT OF A PRODUCT OBTAINED FROM A REACTION.

PERCENT YIELD - IS THE PERCENT OF THE THEORETICAL YIELD OF PRODUCT THAT IS ACTUALLY OBTAINED IN A CHEMICAL REACTION.

- IS THE EFFICIENCY OF A REACTION. IT IS THE RATIO OF THE ACTUAL YIELD TO THE THEORETICAL YIELD.

ACTUAL YIELD

PERCENT YIELD = THERORETICAL YIELD x 100%

9.3E Calculate percent yield, given actual yield and quantity of a reactant. ( )

EXAMPLE

BILLIONS OF POUNDS OF UREA, CO(NH2)2, ARE PRODUCED ANNUALLY FOR USE AS A FERTILIZER. THE REACTION USED IS 2 NH3 + CO2 ---> CO(NH2)2 + H2O

THE TYPICAL REACTION MIXTURE HAS NH3 AND CO2 IN A 3:1 MOL RATIO. IF 47.7 g UREA FORMS PER MOL CO2 THAT REACTS, WHAT IS THE (A) THEORETICAL YIELD, (B) ACTUAL YIELD AND

Solution:

(A) theoretical yield

The stoichiometric mole ratio is 2 mol NH3 : 1 mol CO2

The mole ratio used is 3 mol NH3 : 1 mol CO2

thus CO2 is the limiting reactant

1.00 mol CO2 1 mol CO(NH2)2 60.1 g CO(NH2)2 = 60.1 g CO(NH2)2

1 mol CO2 1 mol CO(NH2)2

(b) actual yield is “47.7 g CO(NH2)2 forms per mol CO2 that reacts”

60.1 g CO(NH2)2