A 0.10 mol sample of each of the four species in the reaction represented above is injected into a rigid, previously evacuated 1.0 L container. Which of the following species will have the highest concentration when the system reaches equilibrium? A) H2S(g) B) CH4(8) C) CS2(8) D) H2(8)

The question is incomplete, complete question is:

A 0.10 mol sample of each of the four species in the reaction represented above is injected into a rigid, previously evacuated 1.0 L container. Which of the following species will have the highest concentration when the system reaches equilibrium?

The equilibrium constant of the reaction =

A) B) C) D)

The correct answer is option A.

Explanation:

The equilibrium constant of the reaction =

0.10 moles of each gas was kept in 1.0 L of container. So. the con concentration of each gas becomes:

Reaction quotient of the reaction :

The reaction will go backward:

Initially

0.10 M  0.10 M        0.10 M     0.10 M

At equilibrium :

(0.10+2x) M  (0.10+x) M        (0.10-x) M     (0.10-4x) M

The expression of an equilibrium constant will be given as';

Solving for x:

x = 0.099

Concentration of species at equilibrium :

The highest concentration when the system reaches at equilibrium will be of hydrogen sulfide.

the density of gold = 19.3 g/cm3

the relation between density and mass and volume is

density = mass / volume

given :

mass of gold = 13 g

density = 13g / volume = 19.3

volume = 13 /19.3 = 0.67 cm^3

thus volume of gold nugget = 0.67 cm^3