Question

A mixture of $254g$ of iodine and $142g$ of chlorine is made to react completely to given a mixture of $ICl$ and $Ic{l}_3$. How many moles of each product are formed?
$(I=127,Cl=35.5)$.

• A. $0.1$ mol of $ICl$ and $0.1$ mol of $IC{l}_3$
• B. $1.0$ mol of $ICl$ and $1.0$ mol of $IC{l}_3$
• C. $0.5$ mol of $ICl$ and $0.1$ mol of $IC{l}_3$
• D. $0.5$ mol of $ICl$ and $1.0$ mol of $IC{l}_3$

Answer 1

The correct option is A $0.1$ mol of $ICl$ and $0.1$ mol of $IC{l}_3$

Given,

Mass of Iodine=$254g$

Mass of Chlorine=$142g$

The reaction involved

$I_2+2C{l}_2⟶ICl+IC{l}_3$

Moles of Iodine provided=$\frac{254g}{254g/mole}=1$ mole

Moles of Chlorine provided=$\frac{142g}{71g/mole}=2$ mole

$\because$ Molar mass of $I_2-254g/mole$

Molar mass of $C{l}_2=71g/mole$

Now, from reaction it is clear that

$1$ mole of $I_2$+$2$ mole of $C{l}_2$ will produce

$1$ mole $ICl$ & $1$ mole $IC{l}_3$

So we are provided with $1$ mole $I_2$ & $2$ mole $C{l}_2$

$\therefore$ Product formed with

$ICl=1$ mole

$IC{l}_3=1$ mole