The following reaction was performed in a sealed vessel at 791 ∘C : H2(g)+I2(g)⇌2HI(g) Initially, only H2 and I2 were present at concentrations of [H2]=3.10M and [I2]=2.50M . The equilibrium concentration of I2 is 0.0800 M . What is the equilibrium constant, Kc, for the reaction at this temperature?

4.31 × 10²

Explanation:

Equation of the reaction;

    ⇌    

The ICE Table is shown as follows:

                                                   ⇌    

Initial                    3.10                     2.50                  0      

Change                 - x                       -x                     + 2x      

Equilibrium        (3.10 - x)                0.0800              2x

From   ;

We can see that 2.50 - x = 0.0800

So; we can solve for x;

x = 2.50 - 0.0800

x = 2.42

 which = (3.10 -x) will be :

= 3.10 - 2.42

= 0.68

= 2x

= 2 (2.42)

= 4.84

430.62

≅ 431

= 4.31 × 10²

the solar system used to have much dust particles option no d

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