Page 24 - Q & A STPM 2022 Chemistry

Chemistry Term 1 STPM Chapter 1 Atoms, Molecules and Stoichiometry
Exam Tips
Exam Tips
Change the concentration of KMnO 4 from g dm to mol dm .
–3
–3

(b) 5Fe + MnO 4 + 8H : 5Fe + Mn + 4H 2 O
3+
2+
–
2+
+
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MV Fe 2+
——–— = 5
–
MV MnO 4
5 × 0.019(31.60)
Concentration of Fe = ——–——–——
2+
25.0
= 0.120 mol dm Term
–3
(c) In the reaction, NH 3 OH reduce Fe ions to Fe ions. The amount of Fe
+
2+
2+
3+
ions formed is determined by titrating with manganate(VII) ions. 1
Hence, in 1 dm of solution, Fe (that has reacted with NH 3 OH )
+
3+
3
= the number of moles of Fe (formed) = 0.120
2+
Molar mass of (NH 3 OH) 2 SO 4 = 164
In 1 dm of solution, the number of moles of (NH 3 OH) 2 SO 4
3
Mass
= ——–——–
Molar mass
5.00
= ——
164
= 0.030
+ 2–
(NH 3 OH) 2 SO 4 : 2NH 3 OH + SO 4
In 1 dm solution, the number of moles of NH 3 OH = 2(0.030) = 0.060
+
3
In 1 dm solution, the number of moles of Fe that has reacted
3
3+
= the number of moles of Fe formed = 0.120
2+
0.060 mol NH 3 OH reacts with 0.120 mol Fe .
3+
+
Hence, 1 mol NH 3 OH reacts with 2 mol Fe .
+
3+
(d) Exam Tips
Exam Tips
Write two separate half-equations for the redox reaction, then add up the
two equations in order to obtain the overall equation.
Fe + e : Fe 2+
3+
–
2NH 3 OH : N 2 O + H 2 O + 6H + 4e –
+
+
Overall: 2NH 3 OH + 4Fe : N 2 O + H 2 O + 4Fe + 6H
2+
+
3+
+
Common Errors
Students must multiply the half-equations with the appropriate coefficients
so that the number of electrons in each half-equation is the same.
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01 STPM Q&A Chem T1 layout.indd 19 4/19/19 9:28 AM