CHE3: MOLECULAR THEORY OF GASES

Gay Luissac’s law of gaseous volumes

It states that when gases react, they do so in volumes which bear a simple whole number ratio to one another and to the volume of the product if gaseous, temperature and pressure remain in constant.

H2(g)      Cl2        2HCl(g)

1 vol     : 1 vol    :   2 vol

1cm3      :   1cm:   2cm3

10cm:  10cm3 :    20cm3

Examples of carbon monoxide gas was mixed with 20cm3 of oxygen gas and the mixture agreed to react in an enclosed versal.

  1. Write equation fo the reaction and the reacting volume ratio

2CO(g)  O2      2CO2

2 vol       : 1 vol   :    2 vol

Determine

  1. The volume of carbon dioxide formed

2vol        :  1vol        :   2 vol

20cm3   :  10cm3    :    20cm3 of carbon dioxide

 

  1. Which of the reactants was in excess and by what volume?

Oxygen was in excess by 10cm3

The total volume of the residue gas and give its composition

Unreacted O2 = 20 – 10 = 10 cm3

CO2 formed = 20cm3

Total residue vol

10  20

30cm3

Composition = 10 cm3 of oxygen and 20 cm3 of carbon dioxide

Avogadro’s law

It states that equal volumes of all gases under the same conditions of temperature and pressure contain the same number of molecules.

H2(g)  Cl2 (g)  2HCl(g)

Mole ratio 1      :     1    :     2

Volume ratio  1 vol: 1 vol: 2 vol: (Gay Luissac’s law)

Molecular ratio 1 molecule :1 molecule : 2 molecules (Avogadro’s equations)

 Qualitative interpretation of reaction equations.

H2(g)  Cl2(g)  2HCl2(g)

Interpretation of equation

1 mole of hydrogen gas reacts with 1 mole of chlorine gas to give two moles of hydrogen chloride gas.

One volume of hydrogen gas reacts with one volume of chlorine gas to give two volumes of hydrogen gas.

Or

1 molecule of hydrogen reacts with one molecule of chlorine to give 2 molecules of hydrogen  chloride.

H2(g)  Cl2(g)   2HCl(g)

In mass

=  = 2

= 2  36.5

= 73 grams

2 grams of hydrogeb gas reacts with 71 grams if chlorine to give 73 grams of hydrogen chloride.

Note;

The total mass of reactants is equal to the total mass of products (law of conservation of mass); matter is neither created nor destroyed in a chemical reaction.

  1. Calculate the mass of sodium chloride formed when excess hydrochloric is added to 24g sodium carbonate.

Step 1 – write a balanced equating of the reaction.

Setp 2 –  highlight two subatnces to be ivolved in the calculation and state units to be used in the calculator and work out the theoretical quantities of the substances to be involved in the calculation.

Procedure to calculator

NaCO3(l)     2HCl(aq)    NaCl(aq)  CO2(g)  H2O

(2 )  (12 )

106g = 117g

1g    =

 

24g =

= 26.49

 

Example 2

  1. Sulphuric acid reacts with copper (II) oxide to give copper (II) sulphate. Calculate the mass of copper (II) oxide required to make 6og of copper (II) sulphate (Cu = 63, O =16, H = 1, S = 32)

H2SO4(aq)  CuO(s)                        CuOSO4(aq)  H2O(l)

Given                  (1 )

79.5                                                     159.5

 

1g

60                                                    60g

= 29.9

Potassium chlorate decomposes on heating to give oxygen and potassium chloride. Calculate the volume of oxygen at s.t.p obtain when 10g of potassium chlorate is heated.

(K ) (Molar gas volume at s.t.p =22.4)

Mass (g) 10g                                       vol at s.t.p

2KCl O3(g)                                         2KCl(g)  3O2(g)

(3 )   given               m

245.2g                                                 given               67.2l

1g                                                        given

10g                                                       given

= 2.74l

  1. Calculate the volume of hydrogen evolved at room temperature when excess hydrochloric is added to 8g of  zinc metal.

 

(Zn= 60, H= 1,Cl = 35.5, molar mass gas volume at room temperature is 24l.)

Mass (g) 80g                                       vol at s.t.p

HCl(aq)  Zn(s)                ZnCl(s)  H(g)

give

130.8                                       24

1g

1.47

Ammonium chloride reacts with sodium hydroxide to give sodium chloride, water and ammonia. Calculate how much ammonium chloride will be required to give 896 cm3 of ammonia measure at s.t.p

(N=14, Cl=35.5, H=1, O =16,Na = 23) (molar gas vol at s.t.p = 22.4dm3)

A compound A contains 52.2% of H, 13% carbon, 34.8% oxygen

  1. Calculate the empirical formula
  2. Molecular formula if formula mass of compound is 46
  3. Equation

C2H6O(g)  3O2(g)   2CO2(g)  3H2O(l)

In order to determine the formula of an oxide of copper, dry hydrogen was passed over heated oxide.

  1. State the precautions you would take in carrying out the experiment giving reason for your answer.

The combustion tube should be titled to prevent water from running back on the hot part the tube which will crack it.

  1. 429g of the oxide were reduced to 0.381g of copper metal determine the formula of the oxide (Cu = 63.5)

Mass of O       = 0.429

=  0.381

= 0.048

 

Combining mass

Cu                               O

2                  :                 1

Cu2O1   = Cu2O

Calculate the number of moles of hydrogen gas produced when 3.27g of zinc metal is reacted with excess hydrochloric acid.

Mass 3.27g                                          1 mole

HCl (aq)  Zn(s)                H2(g)    ZnCl2(s)

gives     (1 mole H2)

3.27                             gives

0.05 moles

Calcium carbonate decomposes on heating write a balanced equation

CaCO3(s)                          CaO(s)   CO2 (g)

gives

100g                                                    44g

1g

10.0g

4.4g.

Mass of the residue.

= 5.6g

Graham’s law of diffusion: it states that

The rate of diffusion of a gas is inversely proportional to the square root of its density

Mathematically

Rate of diffusion

Where   is the density of the gas and  is the sign for proportionality

Considering two gases A and B.

   Note:

light gases diffuse faster than heavier gases

Less dense ases diffuse faster than dense gases.

A VIDEO ABOUT GRAHAM’S LAW OF DIFFUSION

VOLUMETRIC ANALYSIS

It deals with measurements of volumes of solutions by titration method from which the concentration in grammes per litre or moles per litre and relative molecular mass can be determined during titration an indicator is used to make the end point of titration.

Commonly used indicators include methyl ornage and phenolphthalein.

End point is the point when the acid is completed by neutralized by the base. It is indicated by the change of the colour. During titration, experiments are repeated to obtain consistent titre value. The average volume of the solution used id determined by adding two consistent readings and dividing the sum by two . burrete readings are recorded to 2 decimal places e.g 20.00 or 20.15.

Note that the second decimal place is always a zero or a five.

Always record the pipette capacity to one decimal place e.g 25.0cm3 or 20.0cm3.

 Experiment 1

You are provide solutions BA1, and BA2 is 0.1m sulphuric acid solution, H2SO4 and BA2 is sodium hydroxide solution, (NaOH) of the unknown concentration. You are required to determine the concentration of BA2 in moles per litre and grammes per litre.

 Procedure

Pipette 25cm3 or 20cm3 of BA2 into a clean conical flask and add to it 2 drops of phenolpthalyne indicator. Titrate the solution with BA1 from the burette.

The end point is when the colour of the indicator just turns colourless. Repeat the titration until you get consistent titres (results) and record them in the table

Capacity of pipette used = 25.0cm3

exp

Volumes of BA1, to be used to calculate the average volume = 13.60, 13.40

Average volume of BA1 used =

= 13.50cm3

This means that 25.00cm3 of BA2 reacts exactly with 13.50cm3 of BA1

 Calculations

  1. Number of moles of sodium hydroxide used

Volume of alkaline 13.50cm3

Molarity 0.1m

1000cm3 of solution contains 0.1 moles of H2SO4

25.0cm3 of the solution contains

= 0.00135 moles.

Number of moles of the acid used

Equation

2NaOH (aq) + H2SO4(aq)    NaSO4(aq) + 2H2O(l)

2                :       1

25.0cm3 of sodium hydroxide contains

= 0.108 moles

Concentration of the sodium hydroxide in gm/litre

Effect of pollutants on the environment

  1. Carbon monoxide is poisounos to man and other animals
  2. Sulphur dioxide is oxidized to sulphur trioxide by oxygen sulphur trioxide dissolves in rain forming acid rain which is harmful to vegetation.
  3. Nitrogen oxides also lead to acid rain.
  4. Hydrogen sulphide may lead to acid but more directly darken buildings which may have lead paints.
  5. Excess carbon dioxide build up in the atmosphere. The consequences is the warming up of the earth’s surface and lower atmosphere leading to the green house effect.

 Water pollution

Is caused by dissolved substances in water, solid suspensions or oil dispersed in the water. The pollutants include untreated sewage.

Soap and detergents products, petroleum products and fertilizers. The pollutants enter the seas via streams and rivers.

When algae and bacteria feed on pollutants derived from detergents and fertilizers thy flourish thereby reducing oxygen supply in water.

 Manufacture of cement

It contains a mixture calcium aluminate (CACALO2)2 and calcium silicate (CaSiO3). The dry product is ground to a powder and little calcium sulphate is added to slow down setting rate of cement. Water is added forming hydrtated calcium aluminate and silicate.

ASSIGNMENT : MOLECULAR THEORY OF MATTER ASSIGNMENT MARKS : 10  DURATION : 1 week, 3 days

 

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