a 3.00 ml aliquot of 0.001 M NaSCN is diluted to 25.0 mL with 0.2 M Fe(NO3)3 and 0.1 M HNO3.?

Refer to: http://occonline.occ.cccd.edu/online/jmlaux/Kf%20Table%20App.%20D.pdf
Kf for FeSCN²⁺ = 8.9 × 10² = 890

Before the formation of FeSCN²⁺:
Initial concentration of SCN⁻ = (0.001 M) × (3/25) = 0.00012 M
Initial concentration of Fe³⁺ = 0.2 M

Equation for the formation of FeSCN²⁺ complex:
____________ Fe³⁺(aq) ____ + ____ SCN⁻(aq) ____ ⇌ ____ FeSCN²⁺(aq) ______ Kf = 890
Initial: _____0.00012 M __________ 0.2 M ______________ 0 M
Change: ______ -y M _____________ -y M ______________ +y M
At eqm: __(0.00012 - y) M ______ (0.2 - y) M ____________ y M

As 0.00012 > y, we can assume that 0.2 ≫ y, and thus [SCN⁻] at eqm = (0.2 - y) M ≈ 0.2 M
No. of moles of SCN⁻ = (0.2 mol/L) × (25/1000 L) = 0.005 mol