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52.1 mL of aqueous 0.255 M Pb(NO3)2 is mixed with 38.5 mL of 0.415 M NaCl. The equation for the precipitate reaction is: Pb(NO3)2 (aq) 2 NaCl (aq) --> PbCl2 (s) 2 NaNO3 (aq) The concentration of Pb2 ion in the solution is _____ M after the reaction is complete.

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Answer:

0.0585 M

Explanation:

  • Pb(NO₃)₂ (aq) + 2NaCl (aq) → PbCl₂ (s) + 2NaNO₃ (aq)

First we calculate the inital number of moles of each reagent, using the given volumes and concentrations:

  • 0.255 M Pb(NO₃)₂ * 52.1 mL = 13.3 mmol Pb(NO₃)₂
  • 0.415 M NaCl * 38.5 mL = 16.0 mmol NaCl

Then we calculate how many Pb(NO₃)₂ moles reacted with 16.0 mmoles of NaCl, using the stoichiometric coefficients of the reaction:

  • 16.0 mmol NaCl * [tex]\frac{1mmolPb(NO_3)_2}{2mmolNaCl}[/tex] = 8.00 mmol Pb(NO₃)₂

Now we calculate the remaining number of Pb(NO₃)₂ moles after the reaction:

  • 13.3 mmol - 8.00 mmol = 5.30 mmol Pb(NO₃)₂

Finally we divide the number of moles by the final volume to calculate the concentration:

  • 5.30 mmol / (52.1 mL + 38.5 mL) = 0.0585 M

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