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120                                                                                     ICSE Chemistry – 10

                                           EXAMPLE 5.7.  200 mL of CO  is collected at STP when a mixture of acetylene
                                                                      2
                                           and oxygen is ignited. Calculate the volume of acetylene and oxygen at STP in
                                           original mixture
                                                          2C H (g) +  5O (g)  o 4CO (g) +  2H O(l)
                                                                                         2
                                                                          2
                                                               2
                                                            2
                                                                                                     2
                                           Solution: The stoichiometry of the reaction gives
                                                          2C H (g) +  5O (g)  o 4CO (g) +  2H O(l)
                                                            2
                                                                          2
                                                               2
                                                                                         2
                                                                                                     2
                                             Volume ratio:   2 vol      5 vol          4 vol
                                            Volume of CO :                            200 mL
                                                     2
                                           So,
                                                 Volume of C H (g) in the original mixture  =   2   × 200 mL = 100 mL
                                                               2
                                                             2
                                                                                          4
                                                  Volume of oxygen in the original mixture  =   5   × 200 mL = 250 mL
                                                                                          4
                                           EXAMPLE 5.8.  Ammonia is oxidised according to the reaction,
                                                          4NH (g) +  5O (g)  o 4NO(g)         +  6H O(l)
                                                                                                     2
                                                               3
                                                                          2
                                           How many litres of NO are formed when 90 litres of oxygen react with ammonia
                                           at STP?
                                           Solution: The reaction stoichiometry gives,
                                                          4NH (g) +  5O (g)  o 4NO(g)         +  6H O(l)
                                                                                                     2
                                                                          2
                                                               3
                                                Volume ratio:   4 vol      5 vol       4 vol
                                             Volume of reactant:    —      90 L
                                           From the volume ratio,
                                                    Volume of NO formed =   4 vol   × 90 L = 72 L
                                                                           5 vol
                                           EXAMPLE 5.9.  24 mL of marsh gas (CH ) was mixed with 106 mL of oxygen and
                                                                               4
                                           then exploded. On cooling, the volume of the mixture became 82 mL of which
                                           58 mL was unchanged oxygen. Which law does this experiment support? Explain
                                           with calculations.
                                           Solution: The combustion of marsh gas (CH ) is described by the reaction,
                                                                                  4
                                                           CH (g)   + 2O (g)  o CO (g)        +  2H O(l)
                                                              4
                                                                                         2
                                                                          2
                                                                                                     2
                                                Volume ratio:   1 vol      2 vol       1 vol         0
                                               Initial volume:   24 mL      106 mL       —           0
                                             Unreacted volume:   —      58 mL           —
                                           Then,               Volume of oxygen reacted  = 106 mL – 58 mL = 48 mL
                                                        Volume of carbon dioxide formed  = 82 mL – 58 mL   = 24 mL
                                           From the calculation, we are
                                                           V(CH )   :  V(O )     :   V (CO )
                                                                           2
                                                                4
                                                                                           2
                                                            24 mL   :   48 mL    :    24 mL
                                                              1     :     2      :      1
                                                    The volumes of the gaseous reactants and products are in a simple ratio.
                                           Thus, this experiment support the Gay-Lussac’s law of gaseous volumes.
                                           Avogadro’s Law
                                           An Italian physicist, Amedeo Avogadro in 1811 suggested that the smallest
                                           particle in gases (except noble gases) were not atoms but molecules. These
                                           molecules could contain two, three or more atoms.
                                                For example, the number of atoms in one molecule of certain gases
                                           are,
                                              Helium (He)    Oxygen (O )      Nitrogen (N )     Hydrogen (H )
                                                                                          2
                                                                        2
                                                                                                             2
                                                    1              2                 2                 2
                                            Chlorine (Cl )    Ozone (O )  Sulphur dioxide (SO )
                                                         2
                                                                       3
                                                                                              2
                                                    2              3                 3
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