Kc= [3.1*10^-2]^2. (mol/L^2) divide by [8.5*10^-1] [3.1*10^-3]^3 (mol/L)^4​

Kc = 12.58

Explanation:

Kc = [0.229]^2*[0.687]^6/[0.221]^4*[0.5685]^3

Kc = (0.052441)(0.10513)/(0.002385)(0.18373)

Kc = 0.0005513/0.000438

Kc = 12.58

Hope that helps!!

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