A sample of 10 dm3 of polluted air at, When 0.1g of a metal hydride MH was allowed , The element germanium was an important component of transistors

 

PROBLEM:

            A sample of 10 dm3 of polluted air at s.t.p. is passed through lime water so that all the CO₂ present is precipitated as CaCO₃. The mass of CaCO₃ formed is 0.05g. What is the percentage by volume of CO₂ the air sample?

 

SOLUITON:

                        Ca(OH)₂ + CO₂                                                CaCO₃ + H2O

            Number of moles  of CaCO₃ formed = 0.05 / 100

                                                                        = 0.0005 moles.

 

Mole ratio between;

            CaCO₃  :   CO₂

                  1    :      1

Therefore, number of moles of CO₂=0.0005 moles.

            Volume of CO₂ at s.t.p. = 0.0005 x 22.4

                                                   = 0.0112 dm³

%age of CO₂ in the air sample = 0.0112/10 x 100

                                                    = 0.112%

PROBLEM:

            When 0.1g of a metal hydride MH was allowed to react with excess of water, 56 cm³ of hydrogen gas was evolved at s.t.p. Calculate the relative atomic mass of M.

 

SOLUTION:

            MH + H₂O                                           MOH + H₂

 

            Number of moles of hydrogen = 56 / 22400

                                                                = 0.0025 moles

            Mole ratio between;

                                                            H2    :    MH

                                                              1     :      1

            Therefore, number of moles of MH = 0.0025 moles.

                                                            Mass of MH

            Moles of MH =                                                          

                                           Relative formula mass of MH

 

               0.1

0.0025 =                                                        

        Relative formula mass of MH

 

            Relative formula mass of MH = 0.1/ 0.0025

                                                               = 40

                        MH = 40, M + 1 = 40, M = 40 – 1

            Therefore relative atomic mass of M=39

 

PROBLEM:

 

            The element germanium was an important component of transistors. It can be made by heating the ore germanite with hydrogen chloride, distilling off the germanium chloride formed, hydrolyzing  the chloride to  the oxide, and reducing the oxide to the metal. When 1.00g of germanite was treated in this way, the germanium present was completely converted into 0.177g of a chloride containing 33.9% by mass of germanium.

           

            Calculate:

(i)                The empirical formula of the chloride.

(ii)              The oxidation number of germanium in the chloride.

(iii)            The percentage of germanium on germanite.

 

SOLUTION:

(i)                Percentage of mass of chlorine = 100 – 33.9

      = 66.1%

                        Mole ratio of;

                                                            Ge   :   Cl

                                                   33.9/72.6   :   66.1/35.5

                                                              0.467    :    1.860

                                                                       1   :   4

            The empirical formula of the chloride  is GeCl4

(ii)              +4

(iii)            Amount of Ge in 0.177g of GeCl₄ = 0.177 / (72.6 + 35.5 x 4)

                                                                                             = 0.177 / 214.6

         = 8.25 x 10^-4 moles

Mass of Ge in o.177g of GeCl₄ = 72.6 x 8.25 x 10^-4.

                                                                                      = 0.0599 g

Percentage of germanium in germanite = 0.0599 / 1 x 100

         = 5.99%