The secondary valence of `Pt^(4+)` is six. Calculate the number of moles of AgCl participated, when excess of `AgNO_3` solution is added to 2L of 0.1M `PtCl_4. 4NH_3` solution.

Explanation

Secondary valence of `Pt^(4+)` is six and it will be satisfied by four `NH_(3)` and two `CI^(-)` ions. The remaining two `Cl^(-)` ions satisfy only primary valence. `PtCl_(4).4NH_(3)` can be written as `[Pt(NH_(3))_(4)Cl_(2)]Cl_(2)`.
One mole of this complex gives 2 moles of chloride ions in the aqueous solution.
Number of moles of complex present in 2L of 0.1M `PtCl_(4).4NH_(3)=MxxV=0.1xx2=0.2`
Number of moles of chloride ions present in 2L of 0.1M `PtCl_(4).4NH_(3)` solution `= 2 xx0.2 = 0.4`.
Number of moles of AgCl precipitated from 2L of 0.1 M `PtCl_(4).4NH_(3)` is 0.4.