In the process of obtaining lead from PbS, or galena, the galena is "roasted" (heated in the presence of oxygen), so that the following reaction occurs: 2PbS(s)+ 3O₂(g)→ 2PbO(s)+ 2SO₂(g)If 50.0 g of PbS are mixed with 25.0 g of oxygen, how many grams of PbO should form?

Question

Quizplus · Accepted Answer

First, we need to calculate the molar mass of PbS and O2 to determine the moles of each reactant. The molar mass of PbS (207.2 g/mol for Pb + 32.07 g/mol for S) is approximately 239.27 g/mol. The molar mass of O2 is approximately 32.00 g/mol. Next, we calculate the moles of PbS and O2:- Moles of PbS = 50.0 g / 239.27 g/mol ≈ 0.209 moles- Moles of O2 = 25.0 g / 32.00 g/mol ≈ 0.781 molesThe stoichiometry of the reaction is 2 moles of PbS per 3 moles of O2. Therefore, PbS is the limiting reagent because, for complete reaction, it would require only 0.209 * (3/2) = 0.3135 moles of O2, and there is more than this amount of O2 present.Since PbS is the limiting reagent, we use its moles to find out how many moles of PbO will form. According to the reaction, 2 moles of PbS produce 2 moles of PbO. Therefore, 0.209 moles of PbS will produce 0.209 moles of PbO.Finally, we calculate the mass of PbO produced:- Molar mass of PbO = 207.2 g/mol (Pb) + 16.00 g/mol (O) = 223.2 g/mol- Mass of PbO = 0.209 moles * 223.2 g/mol ≈ 46.6 gTherefore, 46.6 grams of PbO should form.

In the Process of Obtaining Lead from PbS, or Galena