Page 212 - Chemistry

0.1 Mol Mg = 0.5 mol O2 = 1200cm3
OR
2mg : O 2
2(24) 24000
x
2.4 / 2(24) = / 240000

X = 2.4 x 24000 = 1200cm3
2(2.4)

17. i) C nH 2n, where n = No. of carbon atoms
ii) 70
iii) C sH 10, CH 3CH=CHCH 2CH 3
OR CH 3CH 2CHCH 2= CH 2

18. i) Fe S O H 2O
20.2 / 56 11.5 / 32 23.0 / 16 45.3 / 18
0.36 / 0.36 0.36 / 0.36 1.44 / 0.36 2.52 / 0.36
1 1 4 7

Empirical formula: FeSO 4 + H 2O

ii) 6.95g = 6.95 / 278 = 0.025
3
 0.05 moles in 250cm = 0.025 x 1000 / 250 = 0.1
Concentration = 6.95 / 278 x 1000 / 250 = 0.1

19. a) Zinc is more reactive// higher reduction potential than copper it will react with//
get oxidized in preference to iron oxygen to form Zinc Oxide coat which protects iron
from rusting
ii) Sacrificial protection or cathodic protection

20. Mole of Mg that reacted = Answer in (c) (ii) x 2
1000 2
= 26 = 0.026 √½
1000
Mass of Mg in the alloy = 0.026 x 24
= 0.624g √½

Mass Cu in the alloy = (1.0 – 0.624)

= 0.376g √½
% of Cu = 0.376 x 100
1.0

= 37.6%√½

21. NH (g) + HNO (g) NH 4NO 3(s)
 1
RMM of NH 4NO 3 = 80
Moles of NH 4NO 3 = 4800 = 60moles
 1
80
RMM of NH 3 = 17
Mass of NH 3 = 60 x 17 = 1020KJ  1

22. From the equation of step 3
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