At a certain temperature, 0.4011 mol of n2 and 1.621 mol of h2 are placed in a 3.50 l container. n2(g)+3h2(g)↽−−⇀2nh3(g) at equilibrium, 0.1001 mol of n2 is present. calculate the equilibrium constant, kc .

The equilibrium constant ≈120.07

Explanation:

Step 1: The balanced equation

N2(g)+3H2(g)↽−−⇀2NH3(g)

This means that for 1 mole of N2 consumed, there is needed 3 moles of H2 to produce 2 moles of NH3

Step 2: Calculate initial moles

Reactants:

 ⇒N2: 0.4011 mol es

 ⇒H2: 1.621 mol es

Products

 ⇒NH3: 0 mol  

Step 3: Calculate number of moles that changed

N2: 0.4011 mol - 0.1001 mol = 0.3010 mol

H2: 3 * 0.3010 mol = 0.9030 mol

NH3: 2 * 0.3010 mol = 0.6020 mol

Step 4: Calculate moles at the equilibrium

N2: 0.1001 moles

 ⇒[N2] = 0.1001moles / 3.5L = 0.0286 M

H2: 1.621 moles -0.9030 moles  = 0.718 moles

 ⇒[H2] = 0.718 moles/ 3.5L  : 0.205 M

NH3: 0.6020 mol es

 ⇒[NH3] =  0.6020 / 3.5= 0.172M

Step 5: Calculate Kc

Kc = [NH3]² / ([N2] [H2]³)

Kc= 0.172²/(0.0286*0.205³)

Kc = 0.029584‬/(0.0286* 0.008615125‬)

Kc  ≈120.07

The Kcis a high value, this shows the position of the equilibrium is to the right.

The equilibrium constant ≈120.07

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