Answer:
[tex]\large \boxed{\text{2.20 g Pb}}[/tex]
Explanation:
They gave us the masses of two reactants and asked us to determine the mass of the product.
This looks like a limiting reactant problem.
1. Assemble the information
We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.
Mᵣ: 239.27 32.00 207.2
2PbS + 3O₂ ⟶ 2Pb + 2SO₃
m/g: 2.54 1.88
2. Calculate the moles of each reactant
[tex]\text{Moles of PbS} = \text{2.54 g PbS } \times \dfrac{\text{1 mol PbS}}{\text{239.27 g PbS}} = \text{0.010 62 mol PbS}\\\\\text{Moles of O}_{2} = \text{1.88 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.058 75 mol O}_{2}[/tex]
3. Calculate the moles of Pb from each reactant
[tex]\textbf{From PbS:}\\\text{Moles of Pb} = \text{0.010 62 mol PbS} \times \dfrac{\text{2 mol Pb}}{\text{2 mol PbS}} = \text{0.010 62 mol Pb}\\\\\textbf{From O}_{2}:\\\text{Moles of Pb} =\text{0.058 75 mol O}_{2} \times \dfrac{\text{2 mol Pb}}{\text{3 mol O}_{2}}= \text{0.039 17 mol Pb}\\\\\text{PbS is the $\textbf{limiting reactant}$ because it gives fewer moles of Pb}[/tex]
4. Calculate the mass of Pb
[tex]\text{ Mass of Pb} = \text{0.010 62 mol Pb} \times \dfrac{\text{207.2 g Pb}}{\text{1 mol Pb}} = \textbf{2.20 g Pb}\\\\\text{The reaction produces $\large \boxed{\textbf{2.20 g Pb}}$}[/tex]
