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.


  1. Using the balanced equation from Question #1, 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 :

  • Determine the molar masses for \(PbS\), \(O_2\), and \(Pb\).
  • Calculate the mass of \(Pb\) produced from each reactant using dimensional analysis.
  • Identify \(PbS\) as the limiting reactant because it produces the smaller amount of product.
  • Round the final theoretical yield to three significant figures, which is \(\boxed{2.20 \text{ g}}\).

Explanation

  1. Problem Analysis and Molar Masses

To determine the amount of lead (Pb) produced, we must first analyze the stoichiometry of the reaction:
\[2PbS + 3O_2 \longrightarrow 2Pb + 2SO_3\]
The given reactants are \(2.54 \text{ g}\) of lead (II) sulfide (\(PbS\)) and \(1.88 \text{ g}\) of oxygen (\(O_2\)). We first identify the molar masses:

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

This is a limiting reactant problem. We must calculate the theoretical yield of \(Pb\) from each reactant to determine which one is fully consumed first.

  1. Stoichiometry Calculations

First, we calculate the amount of \(Pb\) that can be produced from \(2.54 \text{ g}\) of \(PbS\):
\[2.54 \text{ g } PbS \times \frac{1 \text{ mol } PbS}{239.26 \text{ g } PbS} \times \frac{2 \text{ mol } Pb}{2 \text{ mol } PbS} \times \frac{207.2 \text{ g } Pb}{1 \text{ mol } Pb} \approx 2.20 \text{ g } Pb\]
Next, we calculate the amount of \(Pb\) that can be produced from \(1.88 \text{ g}\) of \(O_2\):
\[1.88 \text{ g } O_2 \times \frac{1 \text{ mol } O_2}{32.00 \text{ g } O_2} \times \frac{2 \text{ mol } Pb}{3 \text{ mol } O_2} \times \frac{207.2 \text{ g } Pb}{1 \text{ mol } Pb} \approx 8.12 \text{ g } Pb\]
Since \(PbS\) produces less \(Pb\) (\(2.20 \text{ g}\)) compared to \(O_2\) (\(8.12 \text{ g}\)), \(PbS\) is the limiting reactant. Therefore, the theoretical yield of \(Pb\) is dictated by the amount of \(PbS\) present.

  1. Final Conclusion

Based on the calculations, the limiting reactant is \(PbS\), which allows for the production of \(2.20 \text{ g}\) of \(Pb\). Since the original data \((2.54 \text{ g}\) and \(1.88 \text{ g})\) are given to three significant figures, our final answer must also be expressed to three significant figures. Thus, the mass of lead produced is \(\boxed{2.20 \text{ g}}\).

Examples

In industrial metallurgy, calculating limiting reactants is essential for efficiency. For instance, if you are extracting a valuable metal from ore, knowing exactly how much of a fluxing agent (like oxygen or carbon) is needed prevents wasting expensive materials. By calculating the stoichiometry of the reaction, engineers ensure that the maximum amount of pure metal is recovered while minimizing the environmental impact and cost of excess chemicals. This process is a fundamental application of chemical stoichiometry in everyday industrial practices.