Question

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A mixture contains 40% ferrous oxide, 30% ferric oxide and 30% ferousoferic oxide by mass. What is the percentage of iron content by mass in the mixture?


Answer Given :- 73.83%

Need proper explanation. Thanks!

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Answer 1

Answer:

73.766474 %

Explanation:

Pls refer to the attachment above for the solution.

My ans. is 73.766474% and not 73.83% because I have taken the approximate value of atomic mass of iron ( Fe) . Actual atomic mass of Fe is 55.845u , but for the ease of calculation I have taken it 55.8 . In last step , if you put 55.845 instead of 55.8 & calculate it with a calculator , you will get answer 73.825963 which will get rounded off to 73.83 .

Answer 2

Solution :-

As it is A mixture of

40% Ferrous oxide = FeO

30% Ferric oxide = Fe₂O₃

30% Ferousoferic oxide = FeO.Fe₂O₃

Now as we know that the mass of

Fe = 56 u

O = 16 u

***Please consider Attached for further solution .

Now we have got

In 1 kg of mixture we have 738.584 grams of Fe .

Or 0.738584 kg of Fe

So % of Fe in the mixture

$= \dfrac{\textsf{ Mass of Fe in mixture}}{\textsf{Mass of mixture}} \times 100$

$= \dfrac{ 0.738584}{1} \times 100$

$= 73.8584 \%$

So % of Fe in mixture

$\Huge{\boxed{\sf{= 73.86 \% }}}$