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 6.05 kg of galena with 178 L of oxygen gas at 220°C and 2.00 atm? Lead(II) oxide also forms.

Answer 1

The balanced reaction is

2 PbS + 3 O₂ --------> 2 PbO + SO₂

Given volume of O2, V=178 L, temperature, T=220°C=220+273=493 K, Pressuere, P=2 atm and

Gas constant, R=0.0821 L atm mol^-1 K^-1.

Therefore from Ideal gas equation,

PV=nRT

number moles of O2, n=PV/RT

n=(2 atm x 178 L)/(0.0821 L atm mol^-1 K^-1 x 493 K)

n=8.795 mol O2.

Given mass of Galena, PbS=6.05 kg=6.05 x10^3 g (1kg=1000 g)

Molar mass of PbS=239.3 g/mol

Moles of PbS=mass/molar mass=6.05x10^3 g/239.3 g/mol=25.28 mol PbS.

From the equation the mole ratio between PbS:O2=2:3 or 1:1.5

But here PbS:O2=25.28 : 8.795=1:0.347.

So here O2 is limiting reagent.

Therefore moles of SO2=(1/3) x mole sof O2=(1/3) x 8.795 mol=`2.93 mol.

Now at STP means, P=1 atm, at=25°C=293 K,

Volume of SO2=?

PV=nRT

V=nRT/P=(2.93 molx 0.0821 L atm mol^-1 K^-1 x 293 K)/1 atm

V=70.526 L.

Volume of SO2 ~ 70.53 L.

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