SOLUTION : Suppose that $x$ mole of $ICl$ and $y$ mole of $ICl _{3}$ are formed.
Total iodine $\quad: 127 x+127 y=25 \cdot 4$
Total chlorine : $35 \cdot 5 x+\frac{213 y}{2}=14 \cdot 2$
From (I), $\quad x+y=0 \cdot 2, \quad$ From (II), $\quad x+3 y=0 \cdot 4$
$\therefore \quad 2 y=0 \cdot 2 ; \quad y=0 \cdot 1, x=0 \cdot 1, \quad I C l: I C l_{3}=1: 1$.