H2(g)  +  I2(g)   ⇌   2HI(g)When 46 g of I2 and 1 g of H2 gas heated at equilibrium at 4500C, the equilibrium mixture contained 1.9 g of I2. How many moles of I2 and HI are present at equilibrium ?

Moles of I2 taken = 46254=0.181Moles of H2 taken = 12=0.5Moles of I2 remaining = 1.9254=0.0075Moles of I2 used = 0.181  -  0.0075 = 0.1735Moles of H2 used = 0.1735Moles of H2 remaining = 0.5 - 0.1735 = 0.3265Moles of HI formed = 0.1735 x 2 = 0.347At equilibrium, moles of I2 = 0.0075 molesMoles of HI = 0.347 moles