The balanced equation between in `NaCl and AgNO_(3)`
`underset(58.8g)underset(1"mol")(NaCL)+AgNO_(3) to underset(143.5g)underset(1"mol")(AgCl_(3))+NaNO_(3)`
Let x g of NaCl be present in the mixture.
58.5g NaCl produce= 143.5g AgCl
x g NaCl will produce=`(143.5)/(58.5)xx xgAgCl`
The balanced equation between KCl and `AgNO_(3)` is:
`underset(74.5g)underset(1"mol")(KCl)+AgNO_(3) to underset(143.5g)underset(1"mol")(KCl)(AgCl)+KNO_(3)`
KCl present in the mixture=(3.60-g)
74.5g of Cl produce=143.5g of AgCl
(3.6-x)g of KCl will produce=`(143.5)/(74.5)xx(3.6-x)g of AgCl`
`"Thus", (143.5)/(58.5)x+(143.5)/(74.5)x+(143.5)/(74.5)(3.6x-x)=7.74`
x=1.54
`% "of NaCl"=(1.54)/(3.60)xx100=42.7`
`% "of KCl"=((1.54-3.60))/(3.60)xx100=57.3`