i) What is the name of the process used for the industrial preparation of tetraoxosulphate (VI) acid? (ii) State the catalyst used in (a)(i) (iii) Show by means of bdanced chemical equations only, the industrial preparation of tetraoxosulphate (VI) acid from sulphur (IV) oxide.
(b)(i) Distinguish between dehydration and drying (ii) Explain why concentrated tetraoxosulphate (VI) acid cannot be used to dry ammonia (iii) What is the drying agent for ammonia? (c)(i) Give one example of I. a chloride which is soluble in hot water, II. a trioxocarbonate (IV) which does not decompose on heating, Ill. an amphoteric oxide (ii) List three methods for the preparation of salts (iii) State one method for the recovery of salt from its solution.
(d)(i) State Gay Lussac's law of combining volumes (ii) Consider the reaction represented by the following equation: C\(_2\)H\(_{4(g)}\) + 3O\(_{2(g)}\) \(\to\) 2H\(_2\)O\(_{(g)}\) + 2CO\(_{2(g)}\). What is the volume of oxygen required for the complete combustion of 12.5 cm\(^{3}\) of ethene?
(b)(i) Dehydration is the removeal of the element of water from a compound with change in the chemical composition of the compcu while drying is the removal of associated water from a substance without affecting it chemical composition. (ii) Concentrated H\(_2\)SO\(_4\) reacts with ammonia to form ammonium tetraoxosulphate (VI) (iii) CaO or silica gel.
(c)(i) I. lead chloride (PbCl\(_2\)) II. K\(_2\)CO\(_3\) or Na\(_2\)CO\(_3\) III. ZnO or Al\(_2\)O\(_3\)
(ii) I - Reaction of an acid with base (neutralization reaction) - Reactiction of an acid with a metal - Reaction of an acid with trioxocarbonate (IV) - Double decomposition - Direct combination of its constituent element - A metal with halogens
(iii) - Evaporation - Crystallization
(d)(i) Gay lussac's Law of coming volumes states that when gases react, they do so in volumes which are in simple ratio to one another and to the volumes of the products if gaseous, provided that the temperature and pressure remains constant. (ii) C\(_2\)H\(_{4(g)}\) + 3O\(_{2(g)}\) \(\to\) 2H\(_2\)O + 2Cl\(_{2(g)}\) 1 cm\(^3\) of ethene requires 3cm\(^3\) of oxygen 12.5cm\(^3\) of ethene will require 3 x 12.5cm\(^3\) of oxygen = 37.5cm\(^3\)
