Question

how many moles of h2 gas should be mixed with each mol of I2 vapour in order to convert 90% of it into HI [K=50]

Answer 1

$H_2+I_2<=>2HI$

Let $a$ moles of $H_2$ react with $1mol$ of $I_2$

At equilibrium, after $90\%$ dissociation of $H_2$ $K=\frac{[HI]^2}{[H_2][I_2]}=\frac{(1.8a)^2}{0.1a\times(1-0.9a)}=50$

On solving the above equation we get $a=0.6459$

Hence moles of $H_2$ required$=0.6459mol$