Question

A 1.026 g sample of an iron unknown requires 24.35 mL of 0.0195 M MOA solution to reach the end point. The atomic mass of Fe
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Answer #1

1)

c) H+

2)

Answer

d) 12.9℅

Explanation

5Fe2+ (aq) + MnO4- (aq) + 8H+(aq) --------> 5Fe3+(aq) + Mn2+(aq) + 4H2O(l)

stoichiometrically, 1mole of MnO4 reacts with 5moles of Fe2+

moles of MnO4- consumed = (0.0195mol/1000ml) × 24.35ml = 0.0004748mol

moles of Fe present in the sample = 5 × 0.0004748mol = 0.002374mol

mass of Fe present in the sample = 0.002374mol × 55.85g/mol = 0.1326g

percentage of Fe = (0.1326g/ 1.026g) × 100 = 12.9℅

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