A solution of NaCl (aq) is added slowly to a solution of lead nitrate, Pb (NO 3) 2 (aq), until no further precipitation occurs. The precipitate is collected by filtration, dried, and weighed. A total of 12.11 g CaCl2 (s) is obtained from 200.0 mL of the original solution.

Question

Conrad Harrison

Chemistry

15 May, 21:51

A solution of NaCl (aq) is added slowly to a solution of lead nitrate, Pb (NO 3) 2 (aq), until no further precipitation occurs. The precipitate is collected by filtration, dried, and weighed. A total of 12.11 g CaCl2 (s) is obtained from 200.0 mL of the original solution.

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Hans · Answer 1

0.218 M of Pb (NO3) 2

Explanation:

Equation of the reaction

Pb (NO3) 2 (aq) + 2NaCl (aq) - -> PbCl2 (s) + 2NaNO3 (aq)

1 mole of Pb (NO3) 2 reacts to precipitate 1 mole of PbCl2

Molar mass of PbCl2 = 207 + (35.5*2)

= 278 g/mol

Number of moles of PbCl2 precipitated = mass/molar mass

= 12.11/278

= 0.04356 mol

Since 0.04356 moles of PbCl2 was precipitated, therefore by stoichiometry; 0.04356 moles of Pb (NO3) 2 reacted.

Molarity is defined as the number of moles of solute in 1 liter of solution.

Molarity = number of moles/volumes

= 0.04356/0.2

= 0.218 M