Answered

  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 $\left( SO _3\right)$. Write a balanced chemical equation for this process.



$2 PbS+3 O_2 \longrightarrow 2 Pb+2 SO_3$


  1. Using the balanced equation, how many grams of lead will be produced if 2.54 grams of PbS is burned with 1.88 g of $O _2$? Express your answer to the correct number of significant figures and show all work in the form of dimensional analysis. (Hint: There are two separate stoichiometry equations to solve for this problem.)

Verified Answer :

  • The balanced equation for the reaction is \(2PbS + 3O_2 \longrightarrow 2Pb + 2SO_3\).
  • After calculating molar masses and converting given masses to moles, \(PbS\) is identified as the limiting reactant.
  • Based on the \(1:1\) molar ratio between \(PbS\) and \(Pb\), the reaction produces approximately \(0.01062 \text{ moles}\) of lead.
  • The final mass of lead produced is \(0.01062 \text{ mol} \times 207.2 \text{ g/mol} = \boxed{2.20 \text{ g}}\).

Explanation

  1. Balanced Equation and Molar Masses

First, we determine the balanced chemical equation for the reaction of lead(II) sulfide (\(PbS\)) with oxygen (\(O_2\)) to produce lead (\(Pb\)) and sulfur trioxide (\(SO_3\)):
\[2PbS + 3O_2 \longrightarrow 2Pb + 2SO_3\]
Next, we calculate the molar masses of the reactants using the provided atomic masses: \(Pb \approx 207.2 \text{ g/mol}\), \(S \approx 32.06 \text{ g/mol}\), and \(O \approx 16.00 \text{ g/mol}\).

  • Molar mass of \(PbS = 207.2 + 32.06 = 239.26 \text{ g/mol}\)
  • Molar mass of \(O_2 = 2 \times 16.00 = 32.00 \text{ g/mol}\)
  1. Calculating Moles of Reactants

We calculate the moles of each reactant provided using the given masses of \(2.54 \text{ g}\) for \(PbS\) and \(1.88 \text{ g}\) for \(O_2\):

  • Moles of \(PbS = \frac{2.54 \text{ g}}{239.26 \text{ g/mol}} \approx 0.01062 \text{ mol}\)
  • Moles of \(O_2 = \frac{1.88 \text{ g}}{32.00 \text{ g/mol}} \approx 0.05875 \text{ mol}\)
  1. Identifying the Limiting Reactant

To identify the limiting reactant, we compare the stoichiometric requirements. The balanced equation shows a ratio of \(2 \text{ moles of } PbS\) for every \(3 \text{ moles of } O_2\).

  • Using all \(PbS\), we would need \(0.01062 \text{ mol } PbS \times \left(\frac{3 \text{ mol } O_2}{2 \text{ mol } PbS}\right) = 0.01593 \text{ mol } O_2\).

Since we have \(0.05875 \text{ mol } O_2\), which is much more than the required \(0.01593 \text{ mol}\), \(PbS\) is the limiting reactant. Consequently, \(O_2\) is in excess, and the amount of product is determined solely by the amount of \(PbS\).

  1. Calculating the Final Mass of Lead

Now we calculate the grams of lead (\(Pb\)) produced from the limiting reactant \(PbS\). Based on the stoichiometry of the reaction, \(2 \text{ moles of } PbS\) produce \(2 \text{ moles of } Pb\), giving a \(1:1\) molar ratio.

  • Moles of \(Pb = 0.01062 \text{ mol } PbS \times \left(\frac{2 \text{ mol } Pb}{2 \text{ mol } PbS}\right) = 0.01062 \text{ mol } Pb\)
  • Mass of \(Pb = 0.01062 \text{ mol } Pb \times 207.2 \text{ g/mol} \approx 2.20 \text{ g}\)

Rounding to three significant figures, the final amount of lead produced is \(2.20 \text{ g}\).

Examples

Stoichiometry is fundamental in industrial metallurgy. When refineries process raw ores like lead(II) sulfide, they must calculate the exact amount of oxygen required to ensure the reaction proceeds to completion without wasting resources. By identifying the limiting reactant—the ingredient that runs out first—engineers can predict the exact yield of metal produced, allowing for accurate inventory management and cost control in real-world manufacturing environments.

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