AX2	⇌	A2+	+	2X-
s		s		2s

K_sp = = [A²⁺] [X^– ]² = s × (2s)² = 4s³

$$ \Rightarrow $$ 3.2 $$ \times $$ 10^$$-$$11 = 4s³

$$ \Rightarrow $$ s³ = 8 × 10^–12

$$ \Rightarrow $$ s = 2 × 10^–4 mol L^–1

AgI	⇌	Ag+	+	I-
s		s		s

K_sp = s²

$$ \Rightarrow $$ 1.0 × 10^–16 = s²

$$ \Rightarrow $$ s = 1.0 × 10^–8 mol L^–1

$$ \therefore $$ [Ag⁺] = 1.0 × 10^–8 mol L^–1

Also, in 10^–4 N KI solution,

[I^–1] = (10^–4 + 1.0 × 10^–8) mol L^–1

$$ \Rightarrow $$ [I^–1] = (10^–4) mol L^–1

[As 1.0 × 10^–8 mol L^–1 << 1.0 × 10^–4 mol L^–1]

$$ \therefore $$ K_sp of AgI = [Ag⁺][I^–]

= (1.0 × 10^–8)(10^–4)

= 1.0 × 10^–12 mol L^–1

For any reaction to process in forward direction the reaction quotient (Q) must be less than equilibrium constant K_C.

Q < K_c

Balanced chemical equation for Haber's process is as follows :

3H₂ + N₂ $$ \to $$ 2NH₃

It is given that only 50% of the expected product is formed hence only 10 litre of NH₃ is formed.

Therefore, composition of gaseous mixture at the end is as follows :

N₂ used = 5 litres

N₂ left = 30 L – 5 L = 25 L

H₂ used = 15 litres,

H₂ legt = 30 L – 15 L = 15 L

NH₃ = 10 L