Question

A sample of iron ore contains FeS and non-vollatile inert impurity, only . Roasting of this ore converts all FeS into Fe2​O3​ and a 4% loss in weight was observed. If the mass per cent of FeS in the ore is ′x′, then the value of 11x​ is (Fe=56)​

Answer 1

Answer

Mass of ore =100 g

Mass of FeS=x g

Mass of Impurity =y g

x+y=100

moles of FeS=

88

x

2FeS+

2

x

O

2

→Fe

2

O

3

+SO

2

88

x

2×88

x

Mass of Fe

2

O

3

=

2×88

x

×160 g

=0.909 x g

0.909x+y=96 y (as 4% decreased)

x+y=100

0.091x=4

0.909x+y=96

x=

0.091

4

=43.95 g

≈44 g

%FeS=

100

44

×100=44%

11

44

=4

Answer is 4