Question

Calculate the volume, in mL, of 0.230 M Pb(NO3)2 that must be added to completely react...

Calculate the volume, in mL, of 0.230 M Pb(NO3)2 that must be added to completely react with 34.8 mL of 0.417 M NaCl.

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

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Answer #1

Given

Volume of NaCl V1 = 34.8 ml = 0.0348 L

Molarity of NaCl M1 = 0.417 M (mol/L)

No. of moles of NaCl n1 = V1 * M1 = 0.0348 L * 0.417 mol/L = 0.0145 moles

according to reaction stoichometry 2 moles of NaCl will react with 1 mole of Pb(NO3)2

0.0145 moles of NaCl will react with 0.0145/2 = 7.25 * 10-3 moles of Pb(NO3)2

Molarity of Pb(NO3)2 = 0.230 M (mole/L)

Volume of Pb(NO3)2 = No. of moles of Pb(NO3)2 / Molarity of Pb(NO3)2 = 7.25 * 10-3 moles / 0.230 mol/L

= 0.03155 L = 31.55 ml

Volume of Pb(NO3)2 = 31.55 ml

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