December 05, 2021, 09:53:37 AM
Forum Rules: Read This Before Posting

0 Members and 1 Guest are viewing this topic.

#### Susie_Carlsom

• Regular Member
• Posts: 13
• Mole Snacks: +0/-0
« on: July 13, 2007, 11:36:08 AM »

A finely powdered well mixed sample is a mixture containing NaCl and KCl. A 4.624 gram sample of the mixture was dissolved in 80.00 mL of water and treated with 0.264 molar AgNO3(aq). It re­quired 277.0 mL of the AgNO3(aq) solution to combine with all of the chloride ion present. The percent, by weight, of NaCl in the mixture is therefore ______ %

Can anyone help?

#### Yggdrasil

• Retired Staff
• Sr. Member
• Posts: 3213
• Mole Snacks: +483/-21
• Gender:
• Physical Biochemist
« Reply #1 on: July 13, 2007, 02:42:37 PM »
First start out with a balanced chemical reaction.  What happens when you add the AgNO3 with your mixture?  What does this tell you about the number of moles of one component present in your mixture?

#### DrCMS

• Chemist
• Sr. Member
• Posts: 1278
• Mole Snacks: +208/-81
• Gender:
« Reply #2 on: July 13, 2007, 06:19:10 PM »
You can calculate the Cl- content but that gives you no idea of the Na/K mixture.

#### Yggdrasil

• Retired Staff
• Sr. Member
• Posts: 3213
• Mole Snacks: +483/-21
• Gender:
• Physical Biochemist
« Reply #3 on: July 13, 2007, 06:21:59 PM »
You can calculate the Cl- content but that gives you no idea of the Na/K mixture.

Yes it does.  From the number of moles of Cl, you know the total number of moles of Na and K in the mixture.  Also, since you know that you measured 4.624 g of total mixture, you know the total mass of the Na and K.  This gives you two equations and two unknowns.

#### DrCMS

• Chemist
• Sr. Member
• Posts: 1278
• Mole Snacks: +208/-81
• Gender:
« Reply #4 on: July 13, 2007, 07:19:05 PM »
This gives you two equations and two unknowns.

Go on then work out the the Na/K ratio.

#### enahs

• 16-92-15-68 32-7-53-92-16
• Retired Staff
• Sr. Member
• Posts: 2179
• Mole Snacks: +206/-44
• Gender:
« Reply #5 on: July 13, 2007, 07:40:39 PM »
Quote
Go on then work out the the Na/K ratio.

It is quite simple really.

First, determine the amount of mol's of Cl.

You know that (simplified):

NaCl + KCl  ->  AgCl
In other words, all Cl is used by the Silver.

So you have.

4.626 g - X g NaCl   +       X g  KCl    =    mols Cl = mols AgCl
58.443g/mol                 74.551 g/mol

Once you solve for X you know the amount (in grams) of NaCl (4.626 - X) and KCl (X) in the original mixture, so the ratio is pretty easy from that point on.

#### Susie_Carlsom

• Regular Member
• Posts: 13
• Mole Snacks: +0/-0
« Reply #6 on: July 13, 2007, 07:54:23 PM »
Okay so after following the directions all of you gave me I came up with 65.16% as the answer. Does that sound ok ?

#### enahs

• 16-92-15-68 32-7-53-92-16
• Retired Staff
• Sr. Member
• Posts: 2179
• Mole Snacks: +206/-44
• Gender:
« Reply #7 on: July 13, 2007, 08:03:59 PM »
Looks good to me.

#### Susie_Carlsom

• Regular Member
• Posts: 13
• Mole Snacks: +0/-0
« Reply #8 on: July 13, 2007, 08:13:51 PM »
Thanks sooo Much!!!!

#### enahs

• 16-92-15-68 32-7-53-92-16
• Retired Staff
• Sr. Member
• Posts: 2179
• Mole Snacks: +206/-44
• Gender:
« Reply #9 on: July 13, 2007, 08:33:28 PM »
You do realize what you did though? What those numbers are?

You are just equating the mol's to mol's, and using the molecular weight to get mol's.

#### Susie_Carlsom

• Regular Member
• Posts: 13
• Mole Snacks: +0/-0