Question

Roasting galena [lead(II) sulfide] is an early step in the industrial isolation of lead. How many liters of sulfur dioxide, measured at STP, are produced by the reaction of 4.77 kg of galena with 125 L of oxygen gas at 220°C and 2.00 atm? Lead(II) oxide also forms.

Answer 1

We are given following reaction,

2PbS + 3O₂ → 2PbO +2 SO₂

No. of moles of galena (PbS) present = Mass of galena given / Molar mass of galena

Mass of galena (PbS) given = 4.77 Kg = 4770 g                          (1 Kg = 1000 g)

Molar mass of galena (PbS) = 239.27 g /mol

No. of moles of galena (PbS) present = 4770 g / 239.27 g/mol = 19.93 moles

Using the ideal gas equation calculate the no of moles of oxygen,

PV = nRT

where,

P = Pressure of gas = 2 atm

V = Volume of gas = 125 L

R = Gas constant = 0.0821 L-atm / mol-K

T = Temperature of gas = 220 ⁰C = 493 K

n = (2 atm) *(125 L) / (0.0821 L-atm / mol-K) * (493 K)

n = 6.18 moles

Here oxygen is the limiting reagent as according to coeffecient rule oxygen present is not sufficient,

Moles of galena = 19.93 moles

Also, galena and oxygen react in a ratio of 2:3

Thus, moles of oxygen required = 19.93 * (3/2) = 29.895 moles     (But only 6.18 moles are present)

Now accordiing the coeffecient rule,

Oxygen and sulfur dioxide (SO₂) react in a ratio of 3:2

Thus, moles of SO₂ required = 6.18 * (2/3) moles = 4.12 moles

At STP no of litres of SO₂ produced are,

No. of litres of SO₂ produced (V) = nRT / P                                       (Ideal Gas Equation)

No. of litres of SO₂ produced (V) = (4.12 moles) * (0.0821 L-atm / mol-K) * (273K) / (1 atm)

No. of litres of SO₂ produced (V) = 92.34 L