25.4 gm of iodine and 14.2 gm of chlorine are made to react completely to yield a mixture of ICl and $\text{IC}{l}_3$. Calculate the ratio of moles of ICl and $\text{IC}{l}_3$

The balanced chemical equation is

${\text{I}}_2+2\text{C}{l}_2\to \text{IC}l+\text{IC}{l}_3$

Molar mass of ${\text{I}}_2=$$127\times 2=254\text{\hspace{0.17em}g}$

Molar mass of $\text{C}{l}_2=35.5\times 2=71\text{\hspace{0.17em}g}$

Number of moles of ${\text{I}}_2$ consumed in the reaction is

$n_{{\text{I}}_{\text{2}}}=\frac{{\text{mass\hspace{0.17em}of Iodine,I}}_{\text{2}}}{{\text{Molar\hspace{0.17em}mass of I}}_{\text{2}}}=\frac{25.4}{254}=0.1{\text{\hspace{0.17em}mol\hspace{0.17em}of\hspace{0.17em}I}}_{\text{2}}$

Number of moles of $\text{C}{l}_2$ consumed in the reaction is

$n_{\text{C}{l}_2}=\frac{14.2}{71}=0.2\text{\hspace{0.17em}mol\hspace{0.17em}of\hspace{0.17em}C}{l}_2$

From balanced equation, 1 mole of ${\text{I}}_2$ react with 2 moles of $\text{C}{l}_2$ to give 1 mole of each ICl and $\text{IC}{l}_3$.

As per given data, the ratio of moles of ICl and $\text{IC}{l}_3$ is 1 : 1.