1a) Fe2+¿+MnO4→Fe3 +¿+Mn

2+¿+H 2O ¿¿¿

DEPARTMENT OF TEACHER EDUCATION

BELVEDERE TECHNICAL TEACHERS’ COLLEGE

DEPARTMENT OF NATURAL SCIENCES👉SCIENCESECTION👈

DIPLOMA IN EDUCATION (SECONDARY LEVEL)

NAMES: MAKWANA ADMIRE

CANDIDATURE: 15/090

CLASS: DE1-POST ‘O’

SUBJECT: CHEMISTRY

LECTURER: Ms. MADONDO

TOPIC: REDOX TITRATION

DATE: 27/10/15

INTERNAL MARK:COMMENT: ✍

MODERATOR’S MARK:COMMENT: ✍

EXTERNAL MARK: COMMENT:✍

Fe2+¿→Fe3+¿+e−¿¿¿ ¿……………………….×5

MnO4−¿+8 H +¿+5 e−¿→Mn

2+¿+4 H2O¿¿¿ ¿

5 Fe2+¿→5 Fe3+¿+5 Fe−¿¿¿¿

5MnO4−¿+8 H+¿+5e−¿→Mn

2+¿+4H 2O ¿¿¿ ¿

5 Fe2+¿+MnO4−¿+8H2+¿→ 5Fe3+¿+Mn

2+¿+ 4H 2O¿¿ ¿¿¿

b) 5 moles of Fe2+¿¿ reacted with 1 mole of MnO4−¿ ¿ions

c) number of moles of Fe2+¿¿that react with 25cm3 of 0,04 KMnO4

number of moles=concentration×volume

number onmoles=25

1000×0.04=0.001moles

since the ratio is 1:5 Number of moles of Fe2+¿¿

=0.001 × 5 = 0,005 moles

d) The mass of Fe2+¿¿ which are in 200cm3 of the original solution 20cm3 ¿5×10−3

200cm3 ¿more

0.2

0.02×5×10−3=0.05

Therefore mass =number of moles×molecularmass 5×10−2×56 = 2.8g

1e) percentage by mass of iron in impure iron (ɪɪ) salts. Since 2.8g reacted

2.8g10g

×100

=28%