Question

$25.4g$ of iodine and $14.2g$ of chlorine are made to react completely to yield a mixture of $ICl$ and $IC{l_3}$ . Calculate the number of moles of $ICl$ and $IC{l_3}$ formed.

Answer 1

Hint: Iodine and chlorine both are the halogen and have seven electrons in their outermost shell. When they react with each other, they produce a mixture of iodine chloride and iodine trichloride. The number of moles for any substance is determined by dividing the given weight of the substance with the molar mass of the substance.

Complete answer:
he chemical equation of the reaction that takes place between iodine and chlorine is as follows:
\[{I_2} + 2C{l_2} \to ICl + IC{l_3}\]
In the above reaction, we can clearly see that one mole of the iodine reacts with two moles of chlorine to yield a mixture of one mole of $ICl$ and one mole of $IC{l_3}$.
The molar mass of iodine is = $127g$
The given weight of iodine is = $25.4g$
The molar mass of chlorine is = $71g$
The given weight of chlorine is = $14.2g$
As we know that the number of moles of a substance = $n = \dfrac{w}{{{M_w}}}$
The number of moles of iodine = $n = \dfrac{{25.4}}{{127}} = 0.2moles$
The number of moles of chlorine = $n = \dfrac{{14.2}}{{71}} = 0.2moles$
As we have mentioned above also, one mole of iodine reacts with two moles of chlorine to yield one mole each of $ICl$ and $IC{l_3}$.
Thus, chlorine acts as a limiting reagent.
Hence, $0.2moles$ of chlorine will react with $0.1moles$ of iodine to yield $0.1moles$ of $ICl$ and $0.1moles$ of $IC{l_3}$.

Note:
The limiting reagent in a chemical reaction is a reactant that is totally consumed when the chemical reaction is completed. The amount of product formed is limited by this reagent, since the reaction cannot continue without it.