Question

mixture of FeO and Fe2O3 when heated in air to a constant weight gains
10% in its weight. Find the compositions of the original mixture. [Fe = 56]

Answer 1

Answer:

https://brainly.in/question/1441047

Answer 2

Given:

Mixture of iron oxide and ferric oxide heated in air to a constant weight gains 10% in its weight.

To find:

Compositions of original mixture.

Solution:

The balanced chemical reaction involved is-

2FeO + 0.5O₂ →  ferric oxide

2Fe₃O₄ + 0.5O₂ → 3Fe₂O₃

Lets assume the mass of iron oxide = x gram

                      the mass of ferric oxide= y gram

therefore,   x + y = 100       -(1)

2×27g of iron oxide gives = 160 g of ferric oxide

x g of iron oxide gives = $\left {\frac{{160 \cdot x}}{{44 }}}$ g of ferric oxide

Now, 2×232 g of ferrous oxide gives = 3×160 g of ferric oxide

         y g of ferrous oxide gives = $\left {\frac{{160 \cdot 3 \cdot y}}{{2 \cdot 232 }}}$ g of ferric oxide

Now, by gaining 10% in the weight. weight of mixture becomes 110

now,

$\left {\frac{{160 \cdot x}}{{44 }}}$ + $\left {\frac{{3 \cdot160 \cdot y}}{{2 \cdot 232}}}$ = 110        -(2)

now solving equation (1) and (2) we get the value of x and y which is

x = 5.5gram

y = 94.5gram

so, the composition of iron oxide is 5.5gram

and the composition of Ferric oxide is 94.5gram