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Question

A mixture of FeO and Fe3O4 when heated in air to a constant weight gains 55 in its mass. Find the composition of the initial mixture.


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Solution

Step 1: Assuming the amount of mixture:

  • Let the percent of FeO be x.
  • Let the percent of Fe3O4 be (100-x)

Step 2: The conversion of FeO into Fe2O3 is as follows:

4FeO(s)+O2(g)2Fe2O3(s)

From the equation, 288 g of FeO yields 320 g of Fe2O3

Thus for x g of FeO, the amount of Fe2O3 will be as follows:

320288×xgofFe2O3-----------(i)

Step 3: The conversion of Fe3O4 into Fe2O3 is as follows:

2Fe3O4(s)+12O2(g)3Fe2O3(s)

From the equation, 446 g of Fe3O4 yields 480 g of Fe2O3

So, (100-x) g of Fe2O3 will yield, 480464×(100-x)ofFe2O3--------------(ii)

Step 4: Calculation for percentages:

From equation (i) and (ii),

The total amount of Fe2O3 will be:

TotalamountofFe2O3=320288×xg+480464×(100-x)g

According to the question,

320288×x+480464×(100-x)=105x=20.2

Thus, the percent of FeO be 20.2 and the percent of Fe3O4 be 79.8.


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