The constant for the equilibriumH2+I2↔2HI is given by the following expression:
"K=(HI) ^2\/(H2) (I2)" =50.2.
The amount of moles of iodine is:"n(I2)=m(I2)\/M(I2)" = 63.5(g)/254 (g/mole) = 0.25 moles.
Using expression for the equilibrium constant one can find the amount of moles of HI that are in the equilibrium with given amounts of H2 and I2:
n^2(HI) / (125*0.25) =50.2.
50.2 =n^2(HI)/0.3215
N^2(HI)=50.2*0.3125=15.7
N(HI)=3.96mol.
The mass of the HI can be calculated in the following way by multiplying the amount of moles by the molar mass:
"m(HI)=n(HI)\u00d7M(HI)" = 3.96 (mole) × 128 (g/mole) = 507 g.
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