Areaction mixture in a 3.67 l flask at a certain temperature initially contains a mixture of 0.763 grams of h2 and 96.9 grams of i2. at equilibrium, the flask contains 90.4 grams if hi. calculate the equilibrium constant for this reaction at this temperature.
show your work using an ice table.

H2 + I2 = 2HI

we express Keq as follows:

Keq = [HI]^2 / [H2] [I2]

Where the terms represents the concentrations of the substances involved. THe concentrations are as follows:

H2 = 0.763 ( 1 / 18.02) / 3.67 = 0.0115 M

I2 = 96.9 ( 1 / 253.8) / 3.67 = 0.1040 M

HI (at equilibrium) = 90.4 ( 1 / 127.91) / 3.67 = 0.1956 M 

By the ICE table, we can calculate the equilibrium concentrations,
        H2                       I2                           HI
I      0.0115              0.1040                        0
C      -x                        -x                         +2x
 
E  0.0115-0.0963   0.1040-0.0963       0.1926 ===> x = 0.0963

Keq = (0.1926)^2 / (0.0077) (0.0848)
Keq = 56.81

c is having 4 valence electrons and is bonded to three hydrogens and one sulphur atom, therefore, the number of bond pair is equal to four. as all the valence electrons are used in forming the bonds so, there is no lone pair left on carbon.

s is having 6 valence electrons and is bonded to one other sulfur atom and one carbon atom, therefore, the number of bond pair is equal to 2. as only two valence electrons are used in forming the bonds so, four electrons are left. therefore, two lone pairs are present on s.

considering the question as per the attachment

here we can see that the radon nuclide has 222 as the atomic mass number and 86 as the atomic number.

when it undergoes nuclear fission it gives polonium which has atomic mass number as 218 and atomic number will be 84 along with a helium nucleus which has 4 atomic mass number and 2 as its atomic number.

222 = 218 + 4 (atomic mass number)

86 = ? + 2; therefore ? = 86-2 = 84.

the atomic mass number of polonium will be 84.