Roasting galena (lead(II) sulfide) is a step in the industrial extraction of lea

Question

Roasting galena (lead(II) sulfide) is a step in the industrial extraction of lead. How many liters of sulfur dioxide, measured at STP, are produced by the reaction of 3.75kg of galena with 228L of oxygen gas at 220degree C and 2.0 atm pressure? (lead (II) oxide is the second product of the reaction) A mixture of gaseous disulfur difluoride, dinitrogen tetrafluoride, and sulfur tetrafluoride is placed in an effusion apparatus. Rank the gases in order of increasing effusion rate. Find the ratio of effusion rates of disulfur difluoride and dinitrogen tetrafluoride. If gas X is added and it effuses at 0.935 times the rate of sulfur tetrafluoride, find the molar mass of X.

Explanation / Answer

The reaction of roasting of galena is

4PbS + 6O2 --> 4PbO + 4SO2(sulfur dioxide)

molecular weight of PbS= 239

Moles of PbS in 3.75 kg= 3.75*1000/230= 16.30 moles

moles of oxygen in 228 L at 220 deg.c(220+273.15= 493.15K) and 2 atm pressure

is = PV/RT= 2* 228/(0.0821*493.15) =11.26 moles

Molar ratio ( actual ) of PbS : O2= 16.3 :11.26 = 16.3/11.26 :1 =1.45 :1

molar ratio theoretical PBS :O2= 4: 6= 0.66:1

So PbS is limting and O2 is excess So 4moles of PbS gives 4 moles of SO2

16.3 moles of PbS gives 16.3*4/4= 16.3 moles of SO2

1 mole of any gas at STP occupes 22.4 L

16.3 moles of gas occupes 16.3*22.4L= 365 L

2. Molecular formaula Disulfur diflurode : S2F2, Dintrogen tetrafluoride : N2F4 and Sulfur tetra fluoride : SF4

Molar mass : S2F2= 102, N2F4= 104 and SF4= 108

According to Grahams law of effusion, rate of diffusion is inversly proportional to molecular weight

hence lower the molar mass higher the effusion rate, Hecne the order of increasing effusion rate

SF4<N2F4<S2F2

r =rate of effusion

rS2F2/ rN2F4= sqrt(molar mass of N2F4/ Molar mass of S2F2)= sqrt(104/102)= 1.009

rX/RS2F4= 0.935 = Sqrt( molar mass of S2F4/ Molar mass of X)

0.935= Sqrt(108/M). M = molar mas of X

108/M= 0.935*0.935

M= 108/0.935^2= 123.538 g/mole

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