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Calculate the number of milliliters of 0.666 M Ba(OH)2 required to precipitate all of the Pb2 ions in 163 mL of 0.656 M Pb(NO3)2 solution as Pb(OH)2. The equation for the reaction is:
Pb(NO3)2(aq) Ba(OH)2(aq) Pb(OH)2(s) Ba(NO3)2(aq)

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Answer:

161 mL

Explanation:

  • Pb(NO₃)₂(aq) + Ba(OH)₂(aq) → Pb(OH)₂(s) + Ba(NO₃)₂(aq)

First we calculate how many Pb⁺² moles reacted, using the given concentration and volume of the Pb(NO₃)₂ solution:

  • 163 mL * 0.656 M = 107 mmol Pb(NO₃)₂

As 1 millimol of Pb(NO₃)₂ would react with 1 millimol of Ba(OH)₂, to precipitate 107 mmoles of Pb(NO₃)₂ we would require 107 mmoles of Ba(OH)₂.

Using the number of moles and the concentration we can calculate the required number of milliliters:

  • 0.666 M = 107 mmol / x mL
  • x mL = 161 mL

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