Given the reaction PbO + NaCl + H2O → Pb(OH)Cl + NaOH, how many grams of PbO and NaCl are required to produce 10.0 g of Pb(OH)Cl, assuming excess H2O and complete reaction?

• A. PbO: 8.00 g, NaCl: 4.00 g
• B. PbO: 6.67 g, NaCl: 3.33 g
• C. PbO: 4.50 g, NaCl: 2.25 g
• D. PbO: 10.0 g, NaCl: 5.0 g

Answer: (B)

Verified step by step guidance

1. Write the balanced chemical equation for the reaction: \(\mathrm{PbO + NaCl + H_2O \rightarrow Pb(OH)Cl + NaOH}\). Confirm that the stoichiometric coefficients are all 1, meaning 1 mole of PbO reacts with 1 mole of NaCl to produce 1 mole of Pb(OH)Cl.
2. Calculate the molar mass of the product Pb(OH)Cl by summing the atomic masses of Pb, O, H, and Cl: \(M_{Pb(OH)Cl} = M_{Pb} + M_{O} + M_{H} + M_{Cl}\).
3. Determine the number of moles of Pb(OH)Cl produced from 10.0 g using the formula: \(n_{Pb(OH)Cl} = \frac{\text{mass of } Pb(OH)Cl}{M_{Pb(OH)Cl}}\).
4. Use the stoichiometry of the reaction to find the moles of PbO and NaCl required. Since the mole ratio is 1:1:1, the moles of PbO and NaCl needed are equal to the moles of Pb(OH)Cl produced: \(n_{PbO} = n_{NaCl} = n_{Pb(OH)Cl}\).
5. Calculate the mass of PbO and NaCl required by multiplying their moles by their respective molar masses: \(\text{mass}_{PbO} = n_{PbO} \times M_{PbO}\) and \(\text{mass}_{NaCl} = n_{NaCl} \times M_{NaCl}\). These masses correspond to the amounts needed to produce 10.0 g of Pb(OH)Cl.