Question

$0.1$ mole of $H_2$ and $0.2mole$ of $I_2$ are made to react in order to attain equilibrium $H_2+I_2\rightleftharpoons 2HI$. At equilirbrium $20$% $H_2$ is converted into $HI$, calculate amount of $HI$ formed.

Answer 1

$H_2+I_2\rightleftharpoons 2HI$

$0.1$ $0.2$

at equilibrium 20% $H_2$ is converted into $HI$. therefore, the amount of $HI$ is $=2\times 0.1\times \frac{20}{100}=0.04mol$