A mixture of FeO and Fe_2O_3, when heated in air to
a constant weight, gains 8% in its weight. The
percentage of Fe_20_3, in the given mixture is
(Atomic mass of Fe = 56 amu)
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Let the % of FeO in the mixture be =x
So, % of Fe
3
O
4
in the mixture =(100−x)
FeO on heating is converted into Fe
2
O
3
.
288g
4FeO
+O
2
→
320g
2Fe
2
O
3
288 g of FeO yield =320g of Fe
2
O
3
xg of FeO will yield
=
28
320
xg of Fe
2
O
3
464g
2Fe
3
O
4
+
2
1
O
2
→
480g
3Fe
2
O
3
464 g of Fe
3
O
4
yield =480g of Fe
2
O
3
(100-x) g of Fe
2
O
3
will yield =
464
480
(100−x) of Fe
2
O
3
Total Fe
2
O
3
=
288
320
x+
464
480
(100−x)
According to the question:
288
320
x+
464
480
(100−x)=105
x=20.2
So, percentage of FeO=20.2
and percentage of Fe
3
O
4
=79.8
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