[shama]

the q is: Addition of an execc of AgNO₃ to 0.2012 g sample yielded a mixture of AgCl and AgI that weighted 0.4715g . the precipitate was then heated in a stream of cl₂ to convert AgI to AgCl
2AgI + cl₂ ____ 2 AgCl + I₂
the precipitate was found to weigh 0.3922 g after this treatment,, calculate the percentages of KI and NH₄Cl in the sample...

first I said that Agcl equals 0.3922 ,,, so wt of AgI equals 0.4715-0.3922
but then I thought that AgI is .3922 g coa all AgCl in equation comes from converting AgI to AgCl...
but then I said no coz the 0.3922 grams contain the grams of original Agcl
I really don't understand
please help me

[Borek]

so wt of AgI equals 0.4715-0.3922

No - that's the mass difference between AgI and AgCl. You know that 0.4715 is sum of masses of AgCl and AgI - but you don't know how much moles of each. However, you can calculate how many moles of both together is in the sample using information about AgCl mass afterwards. You also know dependence between number of moles and substance mass. That will let you write two equations with two unknowns - one being amount of AgCl and second being amount of AgI.

Assume there was n_AgCl moles of AgCl and n_AgI moles at the begining - and built your equations around these unknowns.

[shama]

thanks alot for your help.. I'm really pleased and greatfull for all what you explained....