Iron oxide $({F e}_{0.94} O)$ exist. What reaction of iron ions i.e. ${F e}^{2 +}$ and ${F e}^{3 +}$ present in this? Calculate their percentage.

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Solution

Balancing charges in $F e_{0.94} O$ .
Let $x$ be the percentage of $F e^{3 +}$ . $100 - x$ be the percentage of $F e$ .
Total charges sum $= 0$ .
$⟹ 0.94 [\frac{3 (x) + 2 (100 - x)}{100}] - 2 = 0$
$⟹ 0.94 [\frac{3 x + 200 - 2 x}{100}] = 2$
$⟹ \frac{x + 200}{100} = \frac{200}{94}$
$⟹ x + 200 = \frac{20000}{94}$
$⟹ x + 200 = 212.8$
$⟹ x = 12.8$
Percentage of $F e^{+ 3}$ is $12.8$
Percentage of $F e^{+ 2}$ is $100 - 12.8 = 87.2$

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