When 100 " mL of " 0.06 M Fe `(NO_3)_3,50mL ` of `0.2M FeCl_3` and `100mL` of `0.26M` Mg `(NO_3)_2`, are mixed In the final solution…..
`[Fe^(3+)]=` ….
`[NO_3^(ɵ)]=`…..
`[Cl^(ɵ)]=`……
`[Mg^(2+)]=`……

Explanation

(a). 100 " mL of " 0.06 M `Fe(NO_3)_3-=6` " mmol of " `Fe^(3+)+6xx3` " mmol of " `NO_3^(ɵ)`
(b). 50 " mL of " 0.26 `M FeCl_3-=100"m mol of "Fe^(3+)+10xx3m" mol of "Cl^(ɵ)`
(c). `100mL` of 0.26 M `Mg(NO_3)_2-=26"m mol of "mg^(2+)+26xx2"m mol of "NO_3^(ɵ)`
(d). Total`-=16m" mol of "Fe^(3+)+70" m mol of "NO_3^(ɵ)+30"m mol of "Cl^(ɵ)+26"m mol of "Mg^(2+)`
Total volume `=100+50+100=250mL`
`[Fe^(3+)]=(16)/(250)=0.064M`
`[NO_3^(ɵ)]=(70)/(250)=0.28M`
`[Cl^(ɵ)]=(30)/(250)=0.12M`
`[Mg^(2+)]=(26)/(250)=0.104M`