1 Dragonfly 4 wings. Quantity relationships: How much? How many? How many dragonflies wings are required for 4 dragonflies to be able to fly?

SCC-CH110/UCD-CH41C Chapter: 9 Instructor: J.T. P1 Quantity relationships: How much? How many? How many dragonflies’ wings are required for 4 drag...
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SCC-CH110/UCD-CH41C

Chapter: 9

Instructor: J.T.

P1

Quantity relationships: How much? How many? How many dragonflies’ wings are required for 4 dragonflies to be able to fly?

1 Dragonfly



4 wings

0

0

?

SCC-CH110/UCD-CH41C

Same Units

Chapter: 9

Instructor: J.T.

P2

1 DF 4 wings = 5 DFs ? Same Units

SCC-CH110/UCD-CH41C

Chapter: 9

Instructor: J.T.

P3

Quantity Relationships in Chemical Reactions: How to solve stoichiometry problems: What we need to know are two subjects as: 1) A balanced chemical reaction. 2) Relationship between mass and mole. EXAMPLE: How many moles of oxygen are required to burn 2.40 moles of ethane gas? 2 C2H6 (g) + 7 O2 (g)

Given



4 CO2 (g) + 6 H2O (l)

Wanted 2 mol C2H6

2.40 mol C 2H 6

0

7 mol O2

? 0

SCC-CH110/UCD-CH41C

Chapter: 9

2 mol 2.4 mol

Instructor: J.T.

P4

7 mol ?

=

2 mol (C2H6) × ? = 7 mol (O2) × 2.4 mol (C2H6)

7 mol ?= × 2.4 mol 2 mol *********************************************** EXAMPLE: Calculate the number of grams of oxygen that are required to burn 155 g of ethane. 2 C2H6 (g) + 7 O2 (g)



4 CO2 (g) + 6 H2O (l)

Given 2 mol C 2H 6

155 g g 30.06 ( ) mol

155 g C 2H 6

Same Units

Wanted 0

7 mol O2

? 0

Same Units

SCC-CH110/UCD-CH41C

Chapter: 9

2 mol 5.15 mol Same Units

Instructor: J.T.

=

7 mol ? Same Units

7 ? = 5.15 × 2 ? = 18.03 mol (oxygen) Wanted in gram!

mass mole = molar mass molar mass of oxygen = 2 × 16.00 (g/mol) ? = 18.03 mol × 32.00 (g/mol) ? = 577 g O2

P5

SCC-CH110/UCD-CH41C

Chapter: 9

Instructor: J.T.

P6

EXAMPLE: How many milligrams of Nickel Chloride are in a solution if 503 mg of Silver Chloride is precipitated in the reaction of Silver Nitrate and Nickel Chloride solution? 2 AgNO3 (aq) + NiCl2 (aq) → 2 AgCl (s) + Ni(NO3)2 (aq)

Wanted 1 mol NiCl2

Given 2 mol

0

? 0

AgCl

503 mg

mg (AgCl)

AgCl mol (AgCl)

? Same Units ¾ (mg) to (g): 503 mg × (1 g/ 1000 mg) = 0.503 g

SCC-CH110/UCD-CH41C

Chapter: 9

Instructor: J.T.

P7

¾ (mass) to (mole) mole = mass / molar mass mole of AgCl = 0.503 g ÷ 143.35 (g/mol) mole of AgCl = 3.51 × 10-3 mol

1 mol 2 mol = ? 72.1 mol ? = 3.51 × 10 −3 mol AgCl ×

1 mol NiCl2 2 mol AgCl

? = 1.75 × 10-3 mol NiCl2 Wanted in milligram!

mass mole = molar mass molar mass of NiCl2= 129.59 (g/mol) Mass of NiCl2 = 1.75 × 10-3 mol × 129.59 (g/mol) =0.227 g Mass of NiCl2 = 227 mg

SCC-CH110/UCD-CH41C

Chapter: 9

Instructor: J.T.

P8

Gas Stoichiometry at Standard Temperature and Pressure:

One mole of any gas at STP occupies 22.4 L EXAMPLE: What volume of hydrogen, at STP, can be released by 42.7 g of zinc as it reacts with hydrochloric acid?

Zn(s) + 2 HCl (aq) → H2 (g) + ZnCl2 (aq)

Given 1 mol Zn

42.7 g g 65.39 ( ) mol

Wanted 22.4 L 0

? 42.7 g

0

Same Units

1 mol 0.653 mol

=

22.4 L ?

SCC-CH110/UCD-CH41C

Chapter: 9

Instructor: J.T.

P9

EXAMPLE: 1.30L of gaseous ethylene is burned. What volume of oxygen is required if both gas volume are measured at STP? → 2CO2 (g) + 2 H2O(l)

C2H4 (g) + 3O2 (g) Given

Wanted

1 mol C 2H 4

22.4 L

3 mol O2

0

3 × 22.4 L

0

? 1.30 L

22.4 L 1.30 L

0

=

67.2 ?

SCC-CH110/UCD-CH41C

Chapter: 9

Instructor: J.T.

P10

Gas Stoichiometry at nonstandard condition:

In STP condition: T= 273 K, and P =760 torr

P1 V1 T1

=

STP =

P2 V2 T2 nonstandard

In the previous example for the reaction: Zn (s) + 2 HCl (aq) → H2 (g) + ZnCl2 (aq) We calculated the volume of hydrogen as 14.6 L, now let’s change the problem. Instead of measuring the hydrogen volume at STP, let’s use t = 21 °C and P = 748 torr. TK=21+273 =294

760 × 14.6 748 V2 = 2731 294 STP

V2= 16.0 L

=

nonstandard

SCC-CH110/UCD-CH41C

Chapter: 9

Instructor: J.T.

P11

14 transistors

22 resistors

18 capacitors

How many of the above electronic circuits (ec) can you assemble from 22 resistors, 18 capacitors, and 14 transistors?

2

+4



+3

Calculate the theoretical yield of ec:

1 ec 14 T × 2T

= 7 ec

1 ec 22 R × = 5.5 ec ≈ 5 ec 4T 1 ec 18 C × = 6 ec 3C Resistor is the limiting item, because 22 of resistors give 5 ec.

SCC-CH110/UCD-CH41C

Chapter: 9

Instructor: J.T.

P12

Limiting Reactants: Example: The fertilizer ammonium nitrate can be made by direct combination of ammonia with nitric acid. If 74.4 g of ammonia is reacted with 159 g of nitric acid, how many grams of ammonium nitrate can be produced? Also calculate the mass of unreacted reactant that is in excess.

NH3 + HNO3



NH4NO3

Convert the number of grams of each reactant to moles. NH3:

74.4 g = 4.37 mol g 17.03 mol

159 g HNO3: 63.02 g = 2.52 mol mol

1 mol NH 4 NO3 4.37 mol NH 3 × = 4.37 mol NH 4 NO3 1 mol NH 3 1 mol NH 4 NO3 2.52 mol HNO3 × = 2.52 mol NH 4 NO3 1 mol HNO3 ∴ The reactant that yields the smaller amount of product is limiting reactant.

1 mol of NH3 ≡ 1 mol of HNO3 2.52 mol ≡ ? ? =2.52 mol of HNO3 4.37 mol – 2.52 mol = 1.85 mol NH3 left

SCC-CH110/UCD-CH41C

Chapter: 9

Instructor: J.T.

P13

1.85 × 17.03 = 31.51 g NH3 left A mixture of 5.0 g of H2 (g) and 10.0 g of O2(g) is ignited. Water forms according to the following combination reaction: 2H2(g) +O2(g) → 2H2O(g) Which reactant is limiting? How much water will the reaction produce?

SCC-CH110/UCD-CH41C

Chapter: 9

Instructor: J.T.

P14

Percent Yield: The calculated amount of product if is based on the assumption that all of the reactant is converted into product is called the theoretical yield. In laboratory or in industrial production, the actual amount of product isolated from a reaction is usually less than the theoretical yield, and it is called actual yield. General solution:

αA + βB



γC

Given: n gram of A

Given: actual yield Wanted: %yield Step 1: Convert mass of A into the mole of A. Step 2: Calculate the theoretical yield:

γ mol C Number of moles A × α mol A

Step 3: Convert mole of C into the mass of C.

SCC-CH110/UCD-CH41C

Chapter: 9

Instructor: J.T.

P15

Step 4: actual yield %Yield = × 100 theoretical yield

Example: A general chemistry student, preparing copper metal by the reaction of 1.274 g of copper sulfate with zinc metal, obtained a yield of 0.392 g copper. What was the percent yield? To calculate %yield, we need both the theoretical yield of copper and the actual yield. CuSO4 (aq) + Zn(s) → Cu(s) + ZnSO4 (aq) Mass to Mole: 1.274 g CuSO 4 = 7.982 × 10 −3 mol CuSO 4 159.6 g mol

The theoretical yield: 7.982 × 10 −3 mol CuSO 4 ×

1 mol Cu 1 mol CuSO 4

= 7.982 × 10 −3 mol Cu

Mole to Mass: 7.982 × 10 −3 mol Cu × 63.55 g

% Yield

=

mol

= 0.5072 g Cu

actual yield 0.392 g × 100 = × 100 = 77.3 % theoretical yield 0.5072 g

SCC-CH110/UCD-CH41C

Chapter: 9

Instructor: J.T.

Liquid Oxygen

Liquid Hydrogen

Main Engines

P16

SCC-CH110/UCD-CH41C

Chapter: 9

Instructor: J.T.

P17

Thermochemical Equations: H2 (l) + O2 (l) → H2O (g) + energy System

Gives off heat

2 NH3 (g) + energy System

Exothermic -ΔH

→ N2(g) + 3 H2 (g)

Absorbs heat

Endothermic +ΔH

The SI unit for energy: joule (J) How many kilojoule of energy are released when 73.0 g C2H6 burns according to: 2C2H6 (g) +7 O2 (g) → 4 CO2(g) + 6H2O (l)+ 3119KJ 2 mol

C 2H 6

0

3119 KJ

? 73.0 g g 30.05 ( ) mol

73.0 g

0

SCC-CH110/UCD-CH41C

Chapter: 9

Instructor: J.T.

P18

ΔH =2.82 × 103 KJ for the photosynthesis reaction by which plants use energy from the sun to form one mole sugar from carbon dioxide and water. How much energy is required to form 454g (1lb) of simple sugar C6H12 O6?

6 CO2 (g) +6 H2O (g) +

→ C6H12O6 (s) + 6O2 (g)

1 mol

Suga

0

2.82 × 103 KJ

?

454 g g 180 ( ) mol

454 g

0

∴ 7.11 × 103 KJ

SCC-CH110/UCD-CH41C

Chapter: 9

Instructor: J.T.

P19

SCC-CH110/UCD-CH41C

Chapter: 9

Instructor: J.T.

P20

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