Verified Answer :
- The reaction is \(2 PbS + 3 O_2 \longrightarrow 2 Pb + 2 SO_3\).
- Calculate the initial moles: \(n_{PbS} \approx 0.0106 \text{ mol}\) and \(n_{O_2} = 0.05875 \text{ mol}\).
- Determine theoretical yield: \(PbS\) yields \(2.20 \text{ g}\) of \(Pb\), while \(O_2\) yields \(8.12 \text{ g}\) of \(Pb\).
- \(PbS\) is the limiting reactant, thus the amount of lead produced is \(\boxed{2.20 \text{ g}}\).
Explanation
- Analysis of the Reaction and Molar Masses
To determine the amount of lead (Pb) produced from the reaction of lead(II) sulfide (PbS) and oxygen (O₂), we first look at the balanced chemical equation:
\[2 PbS + 3 O_2 \longrightarrow 2 Pb + 2 SO_3\]
The problem provides the initial masses of the reactants: \(2.54 \text{ g}\) of \(PbS\) and \(1.88 \text{ g}\) of \(O_2\). We must identify the limiting reactant by calculating the theoretical yield of lead from each reactant individually.
The molar masses are calculated as:
- Molar mass of \(PbS = 207.2 \text{ g/mol} (Pb) + 32.06 \text{ g/mol} (S) = 239.26 \text{ g/mol}\)
- Molar mass of \(O_2 = 2 \times 16.00 \text{ g/mol} (O) = 32.00 \text{ g/mol}\)
- Atomic mass of \(Pb = 207.2 \text{ g/mol}\)
- Determining Theoretical Yield and Limiting Reactant
First, calculate the moles of each reactant:
- Moles of \(PbS = \frac{2.54 \text{ g}}{239.26 \text{ g/mol}} \approx 0.010616 \text{ mol}\)
- Moles of \(O_2 = \frac{1.88 \text{ g}}{32.00 \text{ g/mol}} = 0.05875 \text{ mol}\)
Next, determine the theoretical yield of \(Pb\) from each reactant using dimensional analysis:
- Yield from \(PbS\):
\[0.010616 \text{ mol } PbS \times \left( \frac{2 \text{ mol } Pb}{2 \text{ mol } PbS} \right) \times 207.2 \text{ g/mol } Pb \approx 2.20 \text{ g } Pb\]
- Yield from \(O_2\):
\[0.05875 \text{ mol } O_2 \times \left( \frac{2 \text{ mol } Pb}{3 \text{ mol } O_2} \right) \times 207.2 \text{ g/mol } Pb \approx 8.12 \text{ g } Pb\]
- Conclusion and Final Result
Comparing the two theoretical yields, the \(PbS\) produces less lead (\(2.20 \text{ g}\)) compared to \(O_2\) (\(8.12 \text{ g}\)). Therefore, \(PbS\) is the limiting reactant, and the actual amount of lead produced is limited by the amount of \(PbS\) available. With three significant figures in both starting masses, the final mass of lead produced is \(2.20 \text{ g}\). The final answer is \(\boxed{2.20 \text{ g}}\).
Examples
In industrial processes, such as metallurgy, understanding limiting reactants is crucial for efficiency and cost management. For instance, if a smelting plant has a specific supply of ore (PbS) and air (O2), engineers must calculate the maximum possible yield of pure metal to optimize the reaction conditions and minimize waste. By knowing which reactant runs out first, they can adjust the input amounts to ensure no material is squandered, making the entire production cycle more sustainable and economically viable.