${\mathrm{H}}_2\left(\mathrm{g}\right) + {\mathrm{I}}_2\left(\mathrm{g}\right) \rightleftharpoons 2\mathrm{HI}\left(\mathrm{g}\right)$

When 46 g of ${\mathrm{I}}_2$ and 1 g of ${\mathrm{H}}_2$ gas heated at equilibrium at ${450}^0\mathrm{C},$ the equilibrium mixture contained 1.9 g of ${\mathrm{I}}_2$. How many moles of ${\mathrm{I}}_2$ and HI are present at equilibrium ?

Moles of ${\mathrm{I}}_2$ taken = $\frac{46}{254}=0.181$

Moles of ${\mathrm{H}}_2$ taken = $\frac{1}{2}=0.5$

Moles of ${\mathrm{I}}_2$ remaining = $\frac{1.9}{254}=0.0075$

Moles of ${\mathrm{I}}_2$ used = 0.181  -  0.0075 = 0.1735

Moles of ${\mathrm{H}}_2$ used = 0.1735

Moles of ${\mathrm{H}}_2$ remaining = 0.5 - 0.1735 = 0.3265

Moles of HI formed = 0.1735 x 2 = 0.347

At equilibrium, moles of ${\mathrm{I}}_2$ = 0.0075 moles

Moles of HI = 0.347 moles