[Susie_Carlsom]

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?

[Yggdrasil]

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]

You can calculate the Cl- content but that gives you no idea of the Na/K mixture.

[Yggdrasil]

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]

This gives you two equations and two unknowns.

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

[enahs]

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]

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]

Looks good to me.

[Susie_Carlsom]

Thanks sooo Much!!!!

[enahs]

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]

ahhhhhhhhhh