December 23, 2024, 08:33:27 AM
Forum Rules: Read This Before Posting


Topic: Calculating the mass of precipitate  (Read 10617 times)

0 Members and 1 Guest are viewing this topic.

Offline natemorse5

  • Very New Member
  • *
  • Posts: 2
  • Mole Snacks: +0/-0
Calculating the mass of precipitate
« on: January 31, 2011, 02:05:40 AM »
I've been working at this problem in every way I can think of, but I am not coming up with a correct answer.

A correct equation I calculated for the problem:  Pb(ClO3)2 + 2NaI = PbI2 + 2NaClO3

The part of the question I am having trouble solving is:  What mass of precipitate will form if 1.50L of the concentrated Pb(ClO3)2 is mixed with 0.650L of 0.190 NaI?  (Assuming the reaction goes into completion)

Offline rabolisk

  • Chemist
  • Full Member
  • *
  • Posts: 494
  • Mole Snacks: +45/-25
Re: Calculating the mass of precipitate
« Reply #1 on: January 31, 2011, 02:43:55 AM »
When I hear that a reaction goes into completion, the first thing that pops into my head is limiting reagent.

Offline natemorse5

  • Very New Member
  • *
  • Posts: 2
  • Mole Snacks: +0/-0
Re: Calculating the mass of precipitate
« Reply #2 on: January 31, 2011, 02:47:47 AM »
My gues is that the 0.650L of 0.190 NaI is the limiting reagent.  If that is the case, how does it apply here?

Offline AWK

  • Retired Staff
  • Sr. Member
  • *
  • Posts: 7976
  • Mole Snacks: +555/-93
  • Gender: Male
Re: Calculating the mass of precipitate
« Reply #3 on: January 31, 2011, 04:29:36 AM »
Now write down a balanced reaction
AWK

Sponsored Links