Question

Lead can be prepared from galena [lead(II) sulfide] by first roasting the galena in oxygen gas to form lead(II) oxide and sulfur dioxide. Heating the metal oxide with more galena forms the molten metal and more sulfur dioxide. (a) Write a balanced equation for each step, including the state of each chemical. (1) (2) (b) Write an overall balanced equation for the process, including the state of each chemical.    (c) How many metric tons of sulfur dioxide form for every metric ton of lead obtained? ____ metric tons

Answer 1

Part a

The balanced reaction of first step

2PbS(s) + 3O2(g) = 2PbO(s) + 2SO2(g)

lead(II) sulfide + oxygen = lead(II) oxide + sulfur dioxide

The balanced reaction of second step

2PbO(s) + PbS(s) = 3Pb(l) + SO2(g)

lead(II) oxide + lead(II) sulfide = molten lead + sulfur dioxide

Part b

overall balanced equation by adding above two reactions

2PbS(s) + 3O2(g) + 2PbO(s) + PbS(s) = 3Pb(l) + SO2(g) + 2PbO(s) + 2SO2(g)

3PbS(s) + 3O2(g) = 3Pb(l) + 3SO2(g)

Divide by 3, we get

PbS(s) + O2(g) = Pb(l) + SO2(g)

Part c

Pb produced = 1 MT x 1000kg/MT = 1000 kg

Moles of Pb produced = mass/molecular weight

= (1000 kg) / (207.2 kg/kmol)

= 4.83 kmol

Moles of SO2 produced = moles of Pb produced = 4.83 kmol

Mass of SO2 produced = moles x molecular weight

= 4.83 kmol x 64.066 kg/kmol

= 309.20 kg x 1MT/1000kg

= 0.309 MT