Question

$25.4$ g of $I_2$ and $14.2$ g of $C{l}_2$ are made to react completely to yield a mixture of $ICl$ and $IC{l}_3$.
Calculate moles of $ICl$ and $IC{l}_3$ formed.

• A. $0.1,0.1$
• B. $0.2,0.2$
• C. $0.1,0.2$
• D. $0.2,0.1$

Answer 1

The correct option is A $0.1,0.1$

The molar masses of iodine and chlorine are $254$ g/mol and $71$ g/mol respectively.
The balanced chemical equation is $I_2+2C{l}_2\to ICl+IC{l}_3$.
$1$ mol of iodine reacts with $2$ moles of chlorine to produce $1$ mol of $ICl$ and $1$ mol of $IC{l}_3$ respectively.
$25.4$ g of iodine corresponds to $0.1$ mol.
$14.2$ g of chlorine corresponds to $0.2$ mol.
$0.1$ mol of iodine will react with $0.2$ moles of chlorine to produce $0.1$ mole of $ICl$ and $0.1$ mol of $IC{l}_3$ respectively.