Chemical Forums

Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: Susie_Carlsom on July 13, 2007, 11:36:08 AM

Title: Help Please?
Post by: Susie_Carlsom on July 13, 2007, 11:36:08 AM: I have this HW question I cannot figure out please help ?

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?
Title: Re: Help Please?
Post by: Yggdrasil 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?
Title: Re: Help Please?
Post by: DrCMS 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.
Title: Re: Help Please?
Post by: Yggdrasil on July 13, 2007, 06:21:59 PM: Quote from: DrCMS 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.

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.
Title: Re: Help Please?
Post by: DrCMS on July 13, 2007, 07:19:05 PM: Quote from: Yggdrasil on July 13, 2007, 06:21:59 PM
This gives you two equations and two unknowns.

Go on then work out the the Na/K ratio.
Title: Re: Help Please?
Post by: enahs 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.
Title: Re: Help Please?
Post by: Susie_Carlsom 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 ?
Title: Re: Help Please?
Post by: enahs on July 13, 2007, 08:03:59 PM: Looks good to me.
Title: Re: Help Please?
Post by: Susie_Carlsom on July 13, 2007, 08:13:51 PM: Thanks sooo Much!!!! ;D
Title: Re: Help Please?
Post by: enahs 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.
Title: Re: Help Please?
Post by: Susie_Carlsom on July 13, 2007, 10:52:09 PM: ahhhhhhhhhh :)