At a certain temperature K = 1.1 X 10^3 L/mol for the reaction given below. Calculate the concentrations of Fe3+, SCN?, and FeSCN2+ at equilibrium if 0.020 mol of Fe(NO3)3 is added to 1.0 L of 0.10 M KSCN. (Neglect any volume change.)Fe3+(aq) + SCN ?(aq) FeSCN2+(aq)Fe3+MSCN- MFeSCN2+ M

Explanation:

Fe⁺³ + SCN⁻ =  FeSCN²⁺

x           x                x

If x mole of Fe⁺³ reacted , unreacted Fe⁺³ = .02 - x

Remaining SCN⁻  unreacted = .1 - x

K_c =   [ FeSCN²⁺] / [Fe⁺³] [ SCN⁻]

1.1 x 10³ =  x  / ( .02 - x ) (.1 - x )

1100 = x / .002 - .12x ( neglecting x² )

x = .0165

Concentration of Fe⁺³ = .02 - x = .0035 M

Concentration of SCN⁻ = .1 - x = .0835 M

Concentration of FeSCN²⁺ = x = .0165M

2b2t

explanation:

2b2t

hey there!

mass = 6.75 g

volume =   9 cm³

therefore:

d = mass / volume

d   = 6.75 / 9

d = 0.7 g/cm³

answer b