[flyza88]

the question is ...the chloride in a 0.12g sample of 95% pure MgCl2 is to be ppted as AgCl .calc the vol of o.1M AgNO3 sol required to precipitate the chloride and give a 10% excess .((the answer given is 26 ml))

first i calculated the  mass of AgCl precipitate ...

95%= [agcl (mgcl2/agcl) /0.12 ] x 100%

agcl = 0.1716g = 1.197x 10*-3 mol

the eq    mgcl2 + 2 AgNO3  = 2 Agcl + Mg(NO3)2

but now i dont understand bout the 10% excess... how should i proceed...

i assume by taking   1.197 x10*-3 + 1.197 x10*-4(excess of the 10%..i guess )

and it is 2 to 2 ratio..

the no of mol of AGNO3 is 1.3167x10*-3mol

ive calculated ...and got 13 ml of agNO3 instead of 26 ml...
                   

[Borek]

What you have missed is MgCl₂ and AgCl₁.

10% excess means add 10% - or multiply by 1.1. So what you did is OK.

You don't need AgCl mass for anything here - go from MgCl₂ mass to MgCl₂ moles first, then from MgCl₂ moles to AgCl moles.

[tetty]

If your MgCl2 is %95 pure, your mgCl2 is 0,12*95/100=0,114 g
95 g/mol MgCl2 contain 71 G/mol Cl- (2*Cl-)
then 0,114 g MgCl2 contain 0,0852 g Cl-
the mol of Cl- is 0,0852/35,5= 0,0024 mol
Ag++ Cl-------AgCl
Ag: Cl= 1:1
so the mol of Ag+ is 0,0024 mol (that is the AgNO3's mol number too).
ıf my AgNO3 solution is 0,1 M and the mol number is 0,0024 mol then the volume of AgNO3 that I need is 0,0024/0,1= 0,024 L= 24 mL
and %10 excess 24+(24*10/100)= 26,4 mL
NOTE: if you add %10 ecsess of AgNO3 solution, your Cl- is precipitated completely.