Verified Answer :
- Convert reactant masses to moles: \(n(PbS) \approx 0.0106\) mol and \(n(O_2) \approx 0.0588\) mol.
- Determine the limiting reactant by comparing theoretical lead yields: PbS produces \(\approx 0.0106\) mol Pb, while \(O_2\) produces \(\approx 0.0392\) mol Pb; therefore, PbS is limiting.
- Calculate the mass of lead: \(0.0106 \text{ mol} \times 207.2 \text{ g/mol} \approx 2.20 \text{ g}\).
- Final answer: \(\boxed{2.20 \text{ g}}\).
Explanation
- Analyzing Reactants and Stoichiometry
To determine the mass of lead (Pb) produced when 2.54 g of lead(II) sulfide (PbS) reacts with 1.88 g of oxygen gas (O\(_2\)), we use the balanced chemical equation:
\[2PbS + 3O_2 \longrightarrow 2Pb + 2SO_3\]
First, we need to determine the molar masses of the reactants involved:
- Molar mass of PbS: \(Pb (207.20 \text{ g/mol}) + S (32.06 \text{ g/mol}) = 239.26 \text{ g/mol}\)
- Molar mass of O\(_2\): \(2 \times 16.00 \text{ g/mol} = 32.00 \text{ g/mol}\)
- Molar mass of Pb: \(207.20 \text{ g/mol}\)
- Determining the Limiting Reactant
Next, we convert the given masses of the reactants into moles using their molar masses:
- Moles of PbS: \(\frac{2.54 \text{ g}}{239.26 \text{ g/mol}} \approx 0.010616 \text{ mol PbS}\)
- Moles of O\(_2\): \(\frac{1.88 \text{ g}}{32.00 \text{ g/mol}} \approx 0.058750 \text{ mol O}_2\)
Using the stoichiometric coefficients from the balanced equation, we calculate the theoretical amount of lead (Pb) each reactant could produce:
- From PbS: \(0.010616 \text{ mol PbS} \times \left(\frac{2 \text{ mol Pb}}{2 \text{ mol PbS}}\right) = 0.010616 \text{ mol Pb}\)
- From O\(_2\): \(0.058750 \text{ mol O}_2 \times \left(\frac{2 \text{ mol Pb}}{3 \text{ mol O}_2}\right) \approx 0.039167 \text{ mol Pb}\)
Comparing these, PbS is the limiting reactant because it produces less lead.
- Calculating the Mass of Lead Produced
Finally, we calculate the mass of Pb produced based on the limiting reactant (PbS):
\[\text{Mass of Pb} = 0.010616 \text{ mol Pb} \times 207.20 \text{ g/mol Pb} \approx 2.1996 \text{ g}\]
Given the input data has three significant figures (2.54 g and 1.88 g), we round our result to three significant figures:
\[\text{Mass of Pb} = \boxed{2.20 \text{ g}}\]
Examples
Stoichiometry is a vital tool in industrial manufacturing, such as producing metals from their ores. By accurately calculating the amounts of raw materials needed—like predicting the amount of lead obtained from lead sulfide—engineers ensure that resources are not wasted and that production yields meet target requirements. This level of precision is essential in fields ranging from chemical engineering to pharmaceuticals, ensuring both economic efficiency and product safety.