Material balance in lead smelting

A) Lead ore concentrate of composition  is roasted by using 300% theoretical air. During roasted all  is oxidized. Determine for 1000kg concentrate.

1) Volume of air at 1 atm and 273 K

2) Amount and composition of roasted product

3) Amount and composition of flue gas

B) The roasted product determined above is smelted in blast furnace using flux as to produce slag having a ratio equal to 35:50:15. The amount of coke is 16% of weight of concentrate. The coke contains ind

4) Amount of each flux
5) Amount of slag


Solution:
Molecular weight

Roasting reactions are





Volume of air = (theoretical amount)x3

Theoretical air can be calculated from the stoichiometric equation.

Volume of air =

Roasted product consists of We have done earlier materials balance

(Refer lectures 17, 18, 19, 21,).

Amount of roasted product =948 kg
Amount of flue gas=59 kg moles