to obtain pure lead

Question

1.     To obtain pure lead, lead (II) sulfide is burned in an atmosphere of pure oxygen. The products of the reaction are lead and sulfur trioxide (SO3). Write a balanced chemical equation for this process. How many grams of lead will be produced if 2.54 grams of PbS is burned with 1.88 g of O2? Express your answer to the correct number of significantfigures. Be sure to show all steps completed to arrive at the answer. (Hint: be sure to work the problem with both PbS and O2).

Answer 1

Hello, Avin,

The unbalanced chemical equation is:

PbS + O₂ = Pb + SO₃

Balancing the lead is easy, but we quickly note that the SO₃ contains an odd number of O atoms, while the source of oxygen, O₂, is diatomic. I'll deal with this by using a temporary shortcut, which will be to use a coefficient of 1.5 for the O₂, so that we, at least mathematically, can obtain the three oxygens we need for the SO₃. (Don't tell the chemists - they can't easily break apart an O₂ whenever they want just a single atom of oxygen). We'll take care of this problem in the next step.

Using this cheat, we get the following balanced equation:

PbS + 1.5O₂ = Pb + SO₃

We can make this equation legal by simply multiplying by the smallest factor that would make all coefficients whole numbers. In this case, multiply each by 2.

2PbS + 3O₂ = 2Pb + 2SO₃

We now have a legally balanced equation.

We see that we need 3 moles of oxygen for every 2 moles of lead sulfide. That's a molar ratio of 3/2 (moles O₂/moles PbS).

Let's determine the number of moles of each in the masses provided. Divide each mass by the molar mass of the compound. For PbS:

2.54g/239.3g/mole = 0.01062 moles PbS

For O₂ the calculation is 1.88g/32g/mole = 0.05875 moles O₂

The molar ratio tells us that, for the oxygen, we need 1.5X the number of moles of PbS. We have much more than that, so PbS is the limiting reagent. We can return some of the oxygen (1.37 grams out of the 1.88 grams we were given!) to the boss and ask for a raise.

We can now assume all of the PbS is consumed. The equation promises we'll get 2 moles Pb for every 2 moles PbS, a 1:1 molar ratio of Pb to PbS.

Assuming all of the 0.0106 moles of PbS reacts to produce the Pb, we'll have 0.0106 mole of lead in our reaction vessel. Convert that to grams lead by multiplying by the molar mass of lead (207.2 grams/mole). I get 2.20 grams. (3 sig figs). We'll also get 0.0106 moles of SO₃, which at 80.06 g/moles, is 0.850 grams.

We consumed 2.54 g of PbS and 0.059 g of O₂ for a total mass of .3.05 grams. The products sum of 3.05 grams. This is what we'd expect from the law of conservation of mass. (Remember that we returned 1.37 grams of the oxygen to the pointy-haired man in the corner office).

The are more questions answered than asked here, but I hope the additional explanation will help clarify the entire process.

Bob