Question

Find out the percentage of $\sf {Fe}^{2+}$ ions in $\sf Fe_{0.93}O$ molecule.

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Answer 1

*Answer:-

We know that in pure iron oxide(FeO),iron and oxygen are present in the ratio 1 : 1

In wustite (Fe0.93O1.00)(Fe0.93O1.00),some of the Fe²+Fe²+ ions are missing and the number of Fe2+Fe2+ ions present in 0.93 instead of 1.

From here we find the number of missing Fe²+Fe²+ ions

• ⇒1.0−0.93=0.07⇒1.0−0.93=0.07

Since each Fe²+Fe²+ ion carries two units of positive charge so the total positiv charge missing

• ⇒0.07×2=0.14⇒0.07×2=0.14

For maintenance of electrical neutrality this much (ie 0.14) positive charge has to be compensated by the presence of Fe3+Fe3+ ions.

If one Fe3+Fe3+ ion replaces one Fe2+Fe2+ ion then there is an increase of one unit positive charge.

So to compensate 0.14 unit positive charge we require 0.14 Fe³+Fe³+ ions to replace Fe²+Fe²+ ion.

So,100Fe²+100Fe²+ ions have Fe³+Fe³+ions=0.140.930.140.93×100

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Total 8.9%

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