Smelting is primarily used to extract a metal from its ore. The smelting of lead

Question

Smelting is primarily used to extract a metal from its ore. The smelting of lead is a two step process. In the first step, lead (ll) sulfide reacts with oxygen to form lead (ll) oxide and gaseous sulfur dioxide. Calculate the standard enthalpy of this reaction, delta^H degree_rxm, using the standard enthalpy of formation values, delta^5 f found. delta H degree_rxn In the second step, the excess lead (ll) sulfide reacts with lead (ll) oxide to produce pure lead and gaseous sulfur dioxide. Determine the amount of pure lead that can be produced from 4.51 g of lead (ll) sulfide and excess lead (ll) oxide, assuming that all of the lead is recovered.

Explanation / Answer

1) 2PbS + 3O2 --------------> 2PbO + 2SO2 (g)

Gfo [PbO(g)] = -219.4 kJ/mol

Hfo [SO2(g)] = -296.8 kJ/mol

Hfo [PbS(s)] = - 100 kJ/mol

Hfo [O2 (g) ] = 0 kJ/mol

Hfo = Hfo(products) - Hfo( reactants)

= 2 x -219.4 + 2 x -296.8 - [2 x -100 + 0]

= -832.4 kJ/mol

Therefore, Horxn = -832.4 kJ/mol

2) PbS + 2PbO --------> 3Pb + SO2

1 mol 3 mol

239 g 207.2 g

4.51 g ?

? = ( 4.51 g/ 239 g) x 207.2 g Pb

= 3.9 g of Pb

Therefore,

amount of lead produced =  3.9 grams

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