To solve the problem of calculating the mass of lead (II) chloride (PbCl₂) that will be precipitated when mixing 200 cm³ of 0.1 M lead (II) trioxonitrate(V) [Pb(NO₃)₂] and 200 cm³ of 0.1 M hydrochloric acid (HCl), we will follow these steps: ### Step 1: Write the balanced chemical equation When lead (II) nitrate reacts with hydrochloric acid, the reaction can be represented as follows: \[ \text{Pb(NO}_3\text{)}_2 (aq) + 2 \text{HCl} (aq) \rightarrow \text{PbCl}_2 (s) + 2 \text{HNO}_3 (aq) \] From the equation, we can see that 1 mole of lead (II) nitrate reacts with 2 moles of hydrochloric acid to produce 1 mole of lead (II) chloride. ### Step 2: Calculate the number of moles of each reactant **For lead (II) nitrate (Pb(NO₃)₂):** - Volume = 200 cm³ = 0.200 L - Concentration = 0.1 M Using the formula for moles: \[ \text{Moles of Pb(NO}_3\text{)}_2 = \text{Concentration} \times \text{Volume} = 0.1 \, \text{mol/L} \times 0.200 \, \text{L} = 0.02 \, \text{mol} \] **For hydrochloric acid (HCl):** - Volume = 200 cm³ = 0.200 L - Concentration = 0.1 M Using the same formula for moles: \[ \text{Moles of HCl} = 0.1 \, \text{mol/L} \times 0.200 \, \text{L} = 0.02 \, \text{mol} \] ### Step 3: Determine the limiting reactant From the balanced equation, we see that 1 mole of Pb(NO₃)₂ reacts with 2 moles of HCl. Therefore, for 0.02 moles of Pb(NO₃)₂, we would need: \[ \text{Moles of HCl required} = 0.02 \, \text{mol Pb(NO}_3\text{)}_2 \times 2 = 0.04 \, \text{mol HCl} \] However, we only have 0.02 moles of HCl available. This means that HCl is the limiting reactant. ### Step 4: Calculate the moles of PbCl₂ produced Since HCl is the limiting reactant, we can determine how much PbCl₂ will be produced based on the moles of HCl available. According to the balanced equation, 2 moles of HCl produce 1 mole of PbCl₂. Therefore, the moles of PbCl₂ produced will be: \[ \text{Moles of PbCl}_2 = \frac{0.02 \, \text{mol HCl}}{2} = 0.01 \, \text{mol PbCl}_2 \] ### Step 5: Calculate the mass of PbCl₂ To find the mass of PbCl₂, we first need to calculate its molar mass: - Molar mass of Pb = 207 g/mol - Molar mass of Cl = 35.5 g/mol Thus, the molar mass of PbCl₂ is: \[ \text{Molar mass of PbCl}_2 = 207 \, \text{g/mol} + 2 \times 35.5 \, \text{g/mol} = 207 \, \text{g/mol} + 71 \, \text{g/mol} = 278 \, \text{g/mol} \] Now, we can calculate the mass of PbCl₂ produced: \[ \text{Mass of PbCl}_2 = \text{Moles} \times \text{Molar mass} = 0.01 \, \text{mol} \times 278 \, \text{g/mol} = 2.78 \, \text{g} \] ### Conclusion The mass of lead (II) chloride that will be precipitated is **2.78 g**. ### Explanation of Other Options - **B. 5.56 g**: This would imply that 0.02 moles of PbCl₂ were produced, which is incorrect since only 0.01 moles can be produced from the limiting reactant (HCl). - **C. 8.34 g**: This value does not correspond to any stoichiometric calculation based on the limiting reactant and is not supported by the balanced equation. - **D. 11.12 g**: This would suggest that 0.04 moles of PbCl₂ were produced, which is impossible given the amount of HCl available. ### Revision Summary - The balanced equation shows the stoichiometry of the reaction between Pb(NO₃)₂ and HCl. - Identify the limiting reactant to determine how much product can be formed. - Calculate moles of reactants and products using concentration and volume. - Use molar mass to convert moles of the product to grams for the final answer.