What is the percentage of Fe^{3+} in Fe_{0.95 | vedclass

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(B) Let the total number of $Fe$ atoms be $0.95$ and $O$ atoms be $1$.Let the number of $Fe^{2+}$ ions be $x$ and the number of $Fe^{3+}$ ions be $(0.

Vedclass · Accepted Answer

$10.53$

Vedclass · Answer

(B) Let the total number of $Fe$ atoms be $0.95$ and $O$ atoms be $1$.Let the number of $Fe^{2+}$ ions be $x$ and the number of $Fe^{3+}$ ions be $(0.95 - x)$.Since the compound is electrically neutral,the total positive charge must equal the total negative charge.$2(x) + 3(0.95 - x) = 2(1)$$2x + 2.85 - 3x = 2$$-x = 2 - 2.85$$x = 0.85$ (number of $Fe^{2+}$ ions).Number of $Fe^{3+}$ ions $= 0.95 - 0.85 = 0.10$.Percentage of $Fe^{3+} = (0.10 / 0.95) 	imes 100 = 10.53\%$.

What is the percentage of $Fe^{3+}$ in $Fe_{0.95}O$?

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