At a particular temperature, K = 4.1 ✕ 10−6 for the following reaction. 2 CO2(g) 2 CO(g) + O2(g) If 2.3 moles of CO2 is initially placed into a 4.9-L vessel, calculate the equilibrium concentrations of all species.

concentration of = 0.0124 = 12.4 ×10⁻³ M

concentration of = 0.0248 = 2.48 ×10⁻² M

concentration of = 0.4442 M

Explanation:

Equation for the reaction:

               ⇄                 +      

Concentration of   =  = 0.469

For our ICE Table; we have:

                                       ⇄                 +      

Initial                 0.469                              0                           0

Change              - 2x                                +2x                      +x

Equilibrium       (0.469-2x)                       2x                         x

K =

K =

Since the value pf K is very small, only little small of  reactant goes into product; so (0.469-2x)² = (0.469)²

x = 0.0124

∴ at equilibrium; concentration of   = 0.0124 = 12.4 ×10⁻³ M

concentration of = 2x  = 2 ( 0.0124)

= 0.0248

= 2.48 ×10⁻² M

concentration of = 0.469-2x

= 0.469-2(0.0124)

= 0.469 - 0.0248

= 0.4442 M

the answer would be a

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2