What is the numerical value of Kc for the following reaction if the equilibrium mixture at 750 °C contains 0.20 M CO and 0.052 M CO2?

FeO(s) + CO(g) ⇌ Fe (s) + CO2 (g)

The answer to your question is    Kc = 0.26

Explanation:

Data

Kc = ?

[CO] = 0.20 M

[CO₂] = 0.052 M

For this problem, do not consider the concentration of the reactants and products in a solid-state.

Formula

           Kc =

Substitution

           Kc =

Simplification and result

           Kc = 0.26

The equilibrium constant Kc is 0.26

Explanation:

Step 1: Data given

The equilibrium mixture contains 0.20 M of CO and 0.052 M of CO2

Step 2: The balanced equation

FeO(s) + CO(g) ⇌ Fe (s) + CO2 (g)

Step 3: Calculate Kc

Pure solids and liquids are not included in the equilibrium constant expression. This is because they do not affect the reactant amount at equilibrium in the reaction, so they are disregarded and kept at 1. Remember that the activity, a, of any solid or liquid in a reaction is equal to 1

Kc = [CO2] / [CO]

⇒with [CO] = 0.20 M

⇒with [CO2] = 0.052 M

Kc = 0.052 / 0.20

Kc = 0.26

The equilibrium constant Kc is 0.26

2b2t

explanation:

2b2t

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