Calculate K from Equilibrium

Question

The equilibrium constant, K_c , for the following reaction is 7.00×10^-5 at 673 K.

NH₄I(s) NH₃(g) + HI(g)

If an equilibrium mixture of the three compounds in a 6.70 L container at 673 K contains 3.32 mol of NH₄I(s) and 0.439 mol of NH₃, the number of moles of HI present is _____ moles.

Answer 1

[NH₄I] = 3.32 mol / 6.70 L = 0.496 M

[NH₃] = 0.439 mol / 6.70 L = 0.0655 M

[HI] = ?

Kc = [NH₃][HI] / [NH₄I] = 7.00x10^-5

7.00x10^-5 = (0.0655)(HI) / 0.496

[HI] = 5.3x10^-4 M = 5.3x10-4 mol/L

moles HI = 5.3x10^-4 mol / L x 6.70 L = 0.00355 moles HI