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℅
A 1.026 g sample of an iron unknown requires 24.35 mL of 0.0195 M MOA solution...
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