Question

In a flask, 2.055 g of FeSO4 is placed in a 250 Erlenmeyer flask to which 75 m 5 mL of 3 M sulfuric acid has been added. The molar mass of FeSO4 is 15199 reaction is shown below: Balanced equation: 5Fe (aq) + MnO4 (a + 8H - 5Fe(n) + Mn() * * Which species is oxidized during the titration? a) Mn04 b) Fe2+ c) H d) Mn2- Which species is reduced during the titration? a) Mn04 b) Fe2+ c) H d) Mn2- How many moles of Fe are initially in the Erlenmeyer flask? b) 0.06764 molesc) 312.2 moles d) 0.01353 moles a) 0.002706 moles

4) how many moles of MnO4- are requiered to reach the endpoint in this titration?

a)0.002706 moles

b)312.2moles

c)0.01353 moles

d)0.06764 moles

Answer 1

5 Fe+2(aq) + MnO4-(aq) + 8 H+(aq) -------------------- 5 Fe+3(aq) + Mn+2(aq) + 4 H2O(l)

5 mol           1 mol

5 Fe+2(aq) + MnO4-(aq) + 8 H+(aq) -------------------- 5 Fe+3(aq) + Mn+2(aq) + 4 H2O(l)

    +2            +7                                                        +3              +2

The oxidation state of Fe is increases from +2 to +3 . so it undergoes oxidation .

Hence , Fe+2 acts as reducing agent

1) Fe+2 is oxidised , The answer is B

The oxidation state of Mn is decreases from +7 to +2. so it undergoes reduction

Hence , MnO4- acts as oxidizing agent

2) MnO4- is reduced , The answer is A

3)

mass of FeSO4 = 2.055 grams

molar mass of FeSO4 = 151.9 g/mol

number of moles of Fe+2 = mass / molar mass = 2.055/151.9 = 0.01353 mol

The answer is D.

4)

5 Fe+2(aq) + MnO4-(aq) + 8 H+(aq) -------------------- 5 Fe+3(aq) + Mn+2(aq) + 4 H2O(l)

5 mol           1 mol

number of moles of Fe+2 = 0.01353 moles

according to equation

5 mole of Fe+2 = 1 mole of MnO4-

0.01353 moles of Fe+2 = ?

                                        = 1 x 0.01353/5 = 0.002706 mol

The answer is 0.002706 moles

The answer is A.