April 26, 2024, 12:03:10 AM
Forum Rules: Read This Before Posting


Topic: calculating the weight  (Read 3231 times)

0 Members and 1 Guest are viewing this topic.

Offline shylight200

  • New Member
  • **
  • Posts: 6
  • Mole Snacks: +0/-1
calculating the weight
« on: September 05, 2007, 08:43:42 PM »
calculate the weight of Na2CO3 needed to react with 25 mL of 0.1 M HCl


can someone please help me get started? thanks

Offline Yggdrasil

  • Retired Staff
  • Sr. Member
  • *
  • Posts: 3215
  • Mole Snacks: +485/-21
  • Gender: Male
  • Physical Biochemist
Re: calculating the weight
« Reply #1 on: September 05, 2007, 08:56:10 PM »
Always start with a balanced chemical reaction.

Offline shylight200

  • New Member
  • **
  • Posts: 6
  • Mole Snacks: +0/-1
Re: calculating the weight
« Reply #2 on: September 05, 2007, 09:00:23 PM »
ok, i wrote the equation and got a 2 to 1 ratio

so i did:
(0.1 mol HCl)/(1 L) * (1 L)/(1000 mL) * 25 mL * (1 Na2CO3)/(2 HCl) * (106 g)/ (1 mol) and i got 0.1324 g of Na2CO3

is this right?

Offline Yggdrasil

  • Retired Staff
  • Sr. Member
  • *
  • Posts: 3215
  • Mole Snacks: +485/-21
  • Gender: Male
  • Physical Biochemist
Re: calculating the weight
« Reply #3 on: September 05, 2007, 11:11:59 PM »
Looks right to me.  Good job.

Sponsored Links