Page 233 - Chemistry

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Moles of Cl 2 used = 0.031 x 3 = 0.047
2
Mark consequently from the equation

11. (a) Cl 2(g) + H 2S (g) HCl (g) + S (s)
(b) Yellow solid particles deposited in the flask  ½
(c) Excess chlorine and hydrogen sulphide gas should not be emitted into the atmosphere
because they are pollutants /harmful
 ½
12. (a) Chlorine gas

(b) (i) Remove traces of hydrogen chloride gas
(ii) Drying agent
13. (a) Fe 3+
(b) It is an oxidizing agent 

(c) 2Fe(OH) 3 (s) Fe 2O 3 (s) + 3H 2O (l)
14. (i) Anhydrous Calcium Chloride (½mks)
(ii) A white ppt is formed
-
HCl gas forms Cl ions solution which react with silver ions to form silver Chloride which is
insoluble OR
Hcl (aq) + AgNO 3 (aq) HNO 3(aq) = AgCl (s)
- +

Cl (aq) + Ag (aq) AgCl (s)

Acids, bases and salts
1. (a) Proton donor/electron acceptor/a substance which when dissolved in water
dissociates/break to hydrogen ions as the only positive ion.

( b) Water/ H 2O
(c) It is a proton donor/electron acceptor
2. (i) Ethylbutanoate
(ii) CH 3CH 2CH 2 C – O – CH 2 – CH 3
(iii) Esters
O
3. (a) Temporary water hardness . This is because hardness is removed by boiling
2+
(b) - Provide Ca ions needed in formation of strong teeth and bones
- Hard water form a layer of carbonate of lead which prevent water coming in contact with
lead which cause poisoning (award 1mk for any one)
 ½
4. Let x be the mass of FeSO4 crystals in saturated solution
 Mass of water = 45 – x  ½
 ½
X g of FeSO 4 dissolves in (45-x)g of water
100x of FeSO 4 dissolves in 100g of water
45 - x
So, solubility is 100x = 15.65
 ½
45 – x
100x = 15.56 (45 – x)
100x + 15.65x = 15.65 x 45
115.65x = 15.65 x 45
x = 15.65 x 45  ½
115.65
= 6.0895
So solubility = 6.09g of FeSO 4 in 100g of water
5 . (a) Ca(HCO 3) 2(aq) CaCO 3(s) + CO 2 + H 2O (l)
or:- Mg(HCO 3) MgCO 3(s) + CO 2(g) + H 2O (s) (award 1mk for any)
heat
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