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A non-stoichiometric iron oxide has the formula Fe_(0.98)O. Iron exists in this oxide as Fe²+ and Fe³+ ions. What percentage of the total iron atoms exists as Fe³+ (give nearest integer)?

1. 4%
2. 6%
3. 8%
4. 10%

Correct answer: 4%

Solution

In Fe_(0.98)O, there is 0.98 mol of Fe per 1 mol of O. Charge on O²- per formula unit = 2. Let the number of Fe³+ ions = x and Fe²+ ions = (0.98 - x). Charge balance: 3x + 2(0.98 - x) = 2 => 3x + 1.96 - 2x = 2 => x = 0.04. Percentage of Fe³+ = (0.04/0.98)*100 ≈ 4.08% ≈ 4%.

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