Question

An ore of iron contains FeS and some non volatile impurity. On roasting this ore converts all FeS to Fe2O3 and an 8 % loss in mass is observed. Calculate mass percent of FeS in ore

Answer 1

Answer:

The rusting of iron ore reaction is given below:

2FeS + 7/2O2 →Fe2O3 + 2SO2

From this it can be concluded that 2 mole of iron ore results in the formation of 1 mole oxide of iron as a rust.

As the molar mass of FeS = 88

The molar mass of Fe2O3 = 160 g

So, from stoichiometry 176 g of ore is converted into the 160 g of oxide or rust.

As there is 4% loss in ore is observed that concludes that 4% of ore get converted into oxide.

The mass of 4% ore will be :

=4100×176= 7.04 g

7.04 g of ore get converted into oxide mass = 160176×7.04 = 6.4 g

So, the mass percentage of the rust is :

= 6.4160×100= 4%