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Chapter 12: Problem 27

How would you prepare \(9.70 \mathrm{~g}\) of \(\mathrm{PbCl}_{2}(\mathrm{~s})\) from a \(0.100 \mathrm{M}\) solution of \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) and a \(0.200 \mathrm{M}\) solution of \(\mathrm{CaCl}_{2}\) ?

Short Answer

Expert verified
To prepare 9.70 g of PbCl₂, determine the moles of PbCl₂ needed using its molar mass (0.0350 moles). Next, find the moles of both reactants required for the reaction using stoichiometry (also 0.0350 moles). Then, calculate the required volumes of both solutions: 350 mL of 0.100 M Pb(NO₃)₂ and 175 mL of 0.200 M CaCl₂. Mix the calculated volumes of the solutions in a reaction vessel, and a white precipitate of PbCl₂ will form. Finally, filter, wash, and dry the precipitate to obtain 9.70 g of PbCl₂.

Step by step solution

01

Convert the mass of PbCl₂ to moles

Divide the given mass of PbCl₂ by its molar mass to find the moles. The molar mass of PbCl₂ = 207.2 g/mol (Pb) + 2 × 35.45 g/mol (Cl) = 277.1 g/mol Moles of PbCl₂ = \(\frac{9.70 g}{277.1 g/mol}\) = 0.0350 moles
02

Use stoichiometry to find the moles of both reactants required

The balanced equation for the reaction is: Pb(NO₃)₂(aq) + CaCl₂(aq) → 2 KNO₃(aq) + PbCl₂(s) From the balanced equation, 1 mole of Pb(NO₃)₂ reacts with 1 mole of CaCl₂ to produce 1 mole of PbCl₂. Therefore, moles of Pb(NO₃)₂ and moles of CaCl₂ required for the reaction are also 0.0350 moles.
03

Calculate the volumes of the given solutions required

Using the molarity formula, we can find the required volumes of both solutions: Molarity (M) = moles/volume(L) For Pb(NO₃)₂ solution: 0.100 M = 0.0350 moles/volume(L) Volume of Pb(NO₃)₂ solution = 0.0350 moles/0.100 M = 0.350 L or 350 mL For CaCl₂ solution: 0.200 M = 0.0350 moles/volume(L) Volume of CaCl₂ solution = 0.0350 moles/0.200 M = 0.175 L or 175 mL
04

Mix the calculated volumes of the solutions

Prepare a reaction vessel to accommodate at least 0.525 L or 525 mL of solution. Pour 350 mL of 0.100 M Pb(NO₃)₂ solution into the vessel, then slowly add 175 mL of 0.200 M CaCl₂ solution while stirring. A white precipitate of PbCl₂ will form, as the reaction between both solutions occurs. Once the volumes of both solutions have been combined and the reaction is complete, filter the precipitate, wash it with distilled water, and allow it to dry to obtain the desired 9.70 g of PbCl₂.

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