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Application of Gay-Lussac’s Law of Combining Volumes

Gay Lussac’s of combining volumes states that gases react in simple ratio with one another and to volumes of the products provided that temperature and pressure remain constant. In this article, you will understand how to apply this law in calculation by studying the following examples below: 1.     2H 2 + O 2 → H 2 O In the reaction above, what volume of hydrogen would be left over when 300cm 3 of oxygen and hydrogen are exploded in a sealed tube?   1cm 3 of oxygen = 2cm 3 of hydrogen 300cm 3 of oxygen = 2 x 300 = 600cm 3 Volume of left over = 1000 – 600 = 400cm 3 2.     Calculate the volume of carbon (II) oxide required to react with 40cm 3 of oxygen. 2CO + O 2 → 2CO 2 1cm 3 of oxygen = 2cm 3 of CO 40cm 3 of oxygen = 2 x 40 = 80cm 3 3.     Calculate the volume of residual gases that would be produced when 100cm 3 of sulphur (IV) oxide reacts with 20cm 3 of oxygen    2SO 2 + O 2 → 2SO 3...
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2023 WAEC Possible Volumetric Analysis

 



B is a solution of Na2CO3. nH2O prepared by dissolving 5.0g of the hydrated salt in 350cm3 of solution. A is a solution containing 0.1 moles of hydrochloric acid per dm3 of solution.

a)      Put A into the burette and titrate 20 or 25cm3 portions of B using methyl orange as an indicator. Record the volume of your pipette. Tabulate your reading and calculate the average volume of acid used.

b)      From your results and information given, calculate the

i)                    Concentration of B in mole per dm3

ii)                   Value of Na2CO3. nH2O

iii)                 Percentage of water of crystallisation

The equation of reaction is

Na2CO3. nH2O(aq) + 2HCl(aq) → 2NaCl(aq) + nH2O(l) + CO2(g)

( C =12, O =16 , Na =23,H=1) 

 

Solution

The volume of the pipette used = 25.00cm3

An indicator used = methyl orange

                                                                    Titration

Burette readings

Rough

1st

2nd

3rd

Final reading (cm3)

24.50

25.10

27.20

25.00

Initial reading (cm3)

0.00

0.00

2.10

0.00

The volume of A used (cm3)

24.50

25.10

25.10

25.00

 

Average volume of A used = 25.1+25.10 + 25.00/3 = 75.2/3 = 25.06cm3

I)                    Concentration of B in mol/dm3

CaVa/CbVb = a/b

Ca = 0.1mole/dm3

Va = 25.06cm3

Cb =  ??

Vb = 25.00cm3

                         .a = 2

                           .b = 1

0.1   x 25.06/Cbx25 = 2/1

50Cb  =  0.1x25.06

Cb  = 0.1x25.06/50  = 0.05mol/dm3

II)                  value of Na2CO3.nH2O

Mass of anhydrous salt = 0.05 x 106 = 5.3g

Mass of hydrated salt = 5/350 x1000/1 = 14.98g

 

Mass concentration of hydrated salt/molar mass of hydrated salt = mass of anhydrous salt/molar mass of anhydrous salt

Mass conc of hydrated salt = 14.The molar mass of hydrated salt = 106 +n18

Mass conc of anhydrous salt = 5.3g

Molar mass of anhydrous salt = 106

 

14.28/106 +n18 = 5.3/106

5.3(106 + n18) = 14.28 x 106

5.3(106 + n18) = 1513.68

106 + n18   = 1513.68/5.3

106 + n18   = 285.6

.n18    = 285.6 – 106

.n18   = 179.6

 .n = 179.6/18

.n =9.98    = Percentagercentage water of crystallization

 

Mass of water of crystallization = 14.28 -5.3 =8.98g

Mass of hydrated salt              = 14.28g

% of crystallization = 8.98/14.28 x 100 = 62.9%

 

 

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