1 Dragonfly 4 wings - Sacramento · PDF fileP V T P V = STP ... Calculate the theoretical...

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?


?4

51 wings

DFsDF

= Same UnitsSame Units


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) → 4 CO2 (g) + 6 H2O (l) Given Wanted

0

2 mol C2H6

7 mol O2

? 0

2.40 molC2H6


?7

4.22 mol

molmol

=

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

molmolmol 4.2

27? ×=

*********************************************** 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 Wanted

? 0

155 g C2H6

)(06.30

155

molg

g

Same Units Same Units

7 mol O2 0

2 molC2H6


?7

15.52 mol

molmol

=

2715.5? ×=

? = 18.03 mol (oxygen)

Same Units Same Units

Wanted in gram!

massmolarmassmole =

molar mass of oxygen = 2 × 16.00 (g/mol)

? = 18.03 mol × 32.00 (g/mol) ? = 577 g O2


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 Given

0

1 mol NiCl2

2 molAgCl

(mg) to (g): 503 mg × (1 g/ 1000 mg) = 0.503 g

? 503 mg mg (AgCl) 0 AgCl

mol (AgCl)

?Same Units


(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

molmolmol1.72

2?

1=

AgClmolNiClmolAgClmol

211051.3? 23 ××= −

? = 1.75 × 10-3 mol NiCl2

Wanted in milligram!

massmolarmassmole =

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


Gas Stoichiometry at Standard Temperature and Pressure:

One mole of any gas at STP occupies 22.4 LEXAMPLE: 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 Wanted

?4.22

653.01 L

molmol

=

0

1 mol 22.4 L Zn

? 042.7 g

)(39.65

7.42

molgg

Same Units


EXAMPLE: 1.30L of gaseous ethylene is burned. What volume of oxygen is required if both gas volume are measured at STP?

C2H4 (g) + 3O2 (g) → 2CO2 (g) + 2 H2O(l) Given Wanted

0

1 mol C2H4

3 mol O2

022.4 L 3 × 22.4 L

?2.67

30.14.22

=LL

? 01.30 L


Gas Stoichiometry at nonstandard condition:

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

2

22

1

11

TVP

TVP

=

STP = 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

294748

2736.14760 2

1

V=

×

STP = nonstandard

V2= 16.0 L


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?

Calculate the theoretical yield of ec:

ececTecR 55.5

4122 ≈=×

ecCecC 6

3118 =×

Resistor is the limiting item, because 22 of resistors give 5 ec.

→2 + 4 + 3

ecTecT 7

2114 =×


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: molmol

gg 37.4

03.17

4.74= HNO3: mol

molgg 52.2

02.63

159=

343

343 37.4

11

37.4 NONHmolNHmol

NONHmolNHmol =×

343

343 52.2

1152.2 NONHmol

HNOmolNONHmolHNOmol =×

∴ 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


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?


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:

AmolCmolAmolesofNumber

αγ

×

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


Step 4:

100% ×=yieldltheoretica

yieldactualYield

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:

434 10982.7

6.159

274.1 CuSOmol

molgCuSOg −×=

The theoretical yield:

CumolCuSOmol

CumolCuSOmol 3

44

3 10982.71

110982.7 −− ×=××

Mole to Mass: Cugmol

gCumol 5072.055.6310982.7 3 =×× −

%3.771005072.0392.0100% =×=×=

gg

yieldltheoreticayieldactualYield

SCC-CH110/UCD-CH41 Instructor: J.T. P16 C Chapter: 9

Liquid Oxygen

Liquid Hydrogen

Main Engines


Thermochemical Equations:

H2 (l) + O2 (l) → H2O (g) + energy

Gives off heat ExothermicSystem -ΔH

2 NH3 (g) + energy → N2(g) + 3 H2 (g)

System 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

0

2 mol C2H6

3119 KJ

?

073.0 g

)(05.30

0.

molgg

73


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)

01 molSuga

2.82 × 103

KJ

?0454 g

)(180

454

molgg

∴ 7.11 × 103 KJ



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1 Dragonfly 4 wings - Sacramento · PDF fileP V T P V = STP ... Calculate the theoretical...

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