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) $\rightarrow$ 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

To solve this problem, we need to use stoichiometry and convert the given mass of PbS and O2 to mass of PbO. First, we need to determine which reactant is the limiting reagent. We can use the balanced equation to do this:
2PbS(s) + 3O2(g) --> 2PbO(s) + 2SO2(g)

From the equation, we can see that it takes 3 moles of O2 to react with 2 moles of PbS. We can use this information to calculate how many moles of O2 are needed to react with 50.0 g of PbS:

50.0 g PbS x (1 mol PbS/239.26 g PbS) x (3 mol O2/2 mol PbS) x (32.00 g O2/1 mol O2) = 63.3 g O2 needed

Since we only have 25.0 g of O2, O2 is the limiting reagent. We can use the amount of O2 to determine how much PbO will be produced:

25.0 g O2 x (1 mol O2/32.00 g O2) x (2 mol PbO/3 mol O2) x (223.20 g PbO/1 mol PbO) = 46.6 g PbO

Therefore, the answer is choice B, 46.6 g.

In the Process of Obtaining Lead from PbS, or Galena →\rightarrow→

In the Process of Obtaining Lead from PbS, or Galena $\rightarrow$