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 35.2 g of PbS is mixed with 15.5 g of oxygen, how many grams of PbO should form?

Question

Quizplus · Accepted Answer

First, determine the limiting reactant. The molar mass of PbS is approximately 239.26 g/mol (207.2 for Pb + 32.06 for S), and the molar mass of O2 is approximately 32.00 g/mol. For PbS: $ \frac{35.2 \, \text{g}}{239.26 \, \text{g/mol}} = 0.147 \, \text{mol} $ of PbS. For O2: $ \frac{15.5 \, \text{g}}{32.00 \, \text{g/mol}} = 0.484 \, \text{mol} $ of O2. The reaction ratio is 2 mol PbS to 3 mol O2, so for 0.147 mol PbS, $ \frac{3}{2} \times 0.147 = 0.2205 $ mol O2 is needed, which is less than what is available. Thus, PbS is the limiting reactant. From the balanced equation, 2 mol of PbS produce 2 mol of PbO. Therefore, 0.147 mol of PbS will produce 0.147 mol of PbO. The molar mass of PbO is approximately 223.2 g/mol (207.2 for Pb + 16.00 for O), so $ 0.147 \, \text{mol} \times 223.2 \, \text{g/mol} = 32.8 \, \text{g} $ of PbO.

In the Process of Obtaining Lead from PbS, or Galena