The question was attempted by most candidates and performance was good. In (a)(i)­(ill), candidates correctly named the process for the industrial preparation of tetraoxosulphate (VI) acid as contact process, stated that vanadium (V) oxide (V 205) was the catalyst employed' and showed by means of balanced chemical equation the preparation oftetraoxosulphate (VI) acid from sulphur (IV) oxide as follows:
2 802(g)       2S03
S03(g) + H2804(l)          H2S2O7(l) H2S2O7(l) +H20(l)  2H2S04(aq)

However, some of the candidates lost marks in (iii) because they did not show the reversibility sign.
In (b )(i)-(ill), most candidates correctly distinguished between dehydration and drying by stating that dehydration is the removal of water from a compound with a change in chemical composition while drying is the removal of associated water from a substance without affecting its chemical composition, explained that concentrated tetraoxosulphate (VI) reacts with ammonia to' form ammonium tetraoxosulphate (VI) was the reason why it is not used to dry ammonia and gave CaO/silica gel as the drying agent for ammonia.

I. Lead chloride (pbCl2);
II. K2CO3/N2C03;
Ill. ZnO/Al2O)/SnO/PbO2

 In (c)(ii) and (iii), candidates correctly listed three methods for the preparation of salts and stated one method for the recovery of salt from its solution from the following:

(ii) Reaction of an acid with a base/neutralization;
Reaction of an acid with a metal;
Reaction of an acid with trioxocarbonate (IV);
Double decomposition/Precipitation;
Direct combination of its constituent element; A metal with halogens.

(iii) Evaporation
Crystallization

In (d)(i), candidate correctly stated Gay Lussac's Law of combining volumes as follows:
When gases react, they do so in volume which are in simple -ratio to one another and to the volume of the products if gaseous provided that temperature and pressure remain constant.
In (d)(ii), candidates. correctly calculated the volume of oxygen for the complete 'combustion of 12.5cm of ethane as follows:

(ii)             C2H4(g) + 3O2(g)           2H2O(g)  +  2CO2(g)

I volume ofethene == 3 volumes of oxygen (mole ratio of ethene to oxygen),
OR
                                     1 cm3 of ethene requires 3cm3 of oxygen
I :.12.5cm3 of ethene will require 3 x 12.5cm3 oxygen.
= 37.5cm3