[nlong8]

Hello, rhis is a discussion problem in my lab,
if anyone could point me in the right direction it would be sublime.

Excess Silver NItrate was added to 14.04 mL of an unknown solution of HCL. The chloride was precipitated as AgCl. The AgCl was then dried and found to have a mass of .2758g. Calculate the Molar concentration of HCL
Thanks
Nate

[plankk]

Start with the reaction equation.

[nlong8]

AgNo3 + HCl --> AgCl + H + N03
hows that?

[UG]

Calculate the amount of moles of AgCl in the 0.2758g sample. Using n(moles)=mass(g)/Molar mass(g mol^-1)
The number of moles of AgCl equals the number of moles of HCl, look at your equation, it is a 1:1 ratio. rearrange this formula to find the concentration: n=c(mol L^-1) x V(L)

[nlong8]

.001924=.0104xmol/L
=.1374 M   
sound right?

[UG]

Yep! Correct. 

-------

.001924=.0104xmol/L
=.1374 M   
sound right?

Except it is 0.01404. I'm sure that was just a typo though.

[nlong8]

ah, you are amazing,
thank you
nate